喵喵喵 XIX

不等式例题

文献 1

4a.

\(\mathrm{1/4+4p(a)/p(a+b)-s(ab/(a+b)2)}\)

化简之后分子是

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &-2& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &-2& \ &2& \ &2& \ &-2& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &1& \ &2& \ &-6& \ &2& \ &1\ & \ & \ \\ \ & \ & \ &-2& \ &2& \ &2& \ &-2\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &-2& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

这个结构很有名，需要背下来：\(\mathrm{p((a-b)2)}\)。\(\square\)

4b.

\(\mathrm{s((b+c-a)/(5a2+4bc))-1/s(a)}\)

化简之后分子是

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &20& \ &25& \ &-90& \ &25& \ &20& \ & \textcolor{lightgray}{0}\ \\ \ &20& \ &-16& \ &36& \ &36& \ &-16& \ &20\ & \ \\ \ & \ &25& \ &36& \ &-168& \ &36& \ &25\ & \ & \ \\ \ & \ & \ &-90& \ &36& \ &36& \ &-90\ & \ & \ & \ \\ \ & \ & \ & \ &25& \ &-16& \ &25\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &20& \ &20\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \\\end{matrix}\right] \]

刚好踩到 Schur 的取等。角上没有东西，扣掉一个 \(\mathrm{s(ab)s(a2(a-b)(a-c))}\)。剩下

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &45& \ &-90& \ &45& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &4& \ &36& \ &36& \ &4& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &45& \ &36& \ &-228& \ &36& \ &45\ & \ & \ \\ \ & \ & \ &-90& \ &36& \ &36& \ &-90\ & \ & \ & \ \\ \ & \ & \ & \ &45& \ &4& \ &45\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

然后由于负项非常少，\(-90\) 可以直接用两边 \(-45\) 带掉，所以去掉 \(\mathrm{45s(a2b2(a-b)2)}\)。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &4& \ &36& \ &36& \ &4& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &36& \ &-228& \ &36& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &36& \ &36& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &4& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

扣掉四倍的 \(\mathrm{p(a)s(a(a-b)(a-c))}\) 就剩下中间 \(\mathrm{40p(a)}\) 倍的

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &1& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &-6& \ &1\ & \ \\ \ & \ &1& \ &1\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ \end{matrix}\right] \]

这个结构也要背下来：\(\mathrm{s(a(b-c)2)}\)。\(\square\)

4c.

\(\mathrm{s((b2c+abc)/(a3+abc)+1)-8}\)

也就是

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ & \textcolor{lightgray}{0}& \ &-6& \ &2& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &1& \ &-4& \ &4& \ &2& \ &-4& \ &2& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &2& \ &2& \ &-6& \ &4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &-6& \ &4& \ &2& \ &-6& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &1& \ & \textcolor{lightgray}{0}& \ &-4& \ &2& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &2& \ &1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

看着最外圈的 \(1\) 很不顺眼，直接拿掉 \(\mathrm{s(a3b4(a-c)2)}\)。剩下的是 \(\mathrm{p(a)}\) 倍的这个：

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ &2& \ & \textcolor{lightgray}{0}& \ &-4& \ &2& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &-4& \ &4& \ &1& \ &-4& \ &2\ & \ \\ \ & \ &2& \ &1& \ &-6& \ &4& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &-4& \ &4& \ &1& \ &-4\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &-4& \ &2\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &2& \ &1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

先拿掉两个 \(\mathrm{p(a-b)2}\)，然后顶上再拿掉一个 \(\mathrm{s(a3b(a-b)2)}\) 变对称，变成：

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ &-1& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-3& \ & \textcolor{lightgray}{0}& \ &1\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-3& \ &6& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &-1& \ & \textcolor{lightgray}{0}& \ &-3& \ &-1\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

顶上还可以拿一个 \(\frac12\mathrm{s(ab(a2-b2)2)}\)，这样最外圈的 \(-1\) 就没了。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \frac{1}{2}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \frac{1}{2}& \ & \textcolor{lightgray}{0}\ \\ \ & \frac{1}{2}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-3& \ & \textcolor{lightgray}{0}& \ & \frac{1}{2}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-3& \ &6& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-3& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \frac{1}{2}& \ & \frac{1}{2}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

第二行继续拿，减掉最强的结构 \(\mathrm{s(c(a+b)(a-b)4)}\)（当然 \(\times\frac12\)）。剩下的是 \(\mathrm{p(a)}\) 倍的

\[\left[\begin{matrix} 3& \ &-1& \ &-4& \ &3\ \\ \ &-4& \ &6& \ &-1\ & \ \\ \ & \ &-1& \ &-4\ & \ & \ \\ \ & \ & \ &3\ & \ & \ & \ \end{matrix}\right] \]

拿掉两个 \(\mathrm{Schur_3}\) 就可以用 \(\mathrm{s(a(a-c)2)}\) 带过了。

也就是说

\[\mathrm{LHS=s(a3b4(a-c)2)+2p(a)p((a-b)2)+p(a)s(a3b(a-b)2)}\\ \kern{40pt}\mathrm{+1/2p(a)s(ab(a2-b2)2)+1/2p(a)s(c(a+b)(a-b)4)}\\ \mathrm{+2p(a)2s(a(a-b)(a-c)+p(a)2s(a(a-c)2)} \]

\(\square\)

4c.

\(\mathrm{s(a5)2-3p(a)s(a7)}\)

次数太高，暂时先留在这。

4d.

\(\mathrm{3s((a+b)/(a2+ab+b2))-16s(ab)/p(a+b)}\)

也就是

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &3& \ &-4& \ &2& \ &-4& \ &3& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &6& \ &-2& \ &-4& \ &-4& \ &-2& \ &6& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &3& \ &-2& \ &4& \ &2& \ &4& \ &-2& \ &3\ & \ & \ \\ \ & \ & \ &-4& \ &-4& \ &2& \ &2& \ &-4& \ &-4\ & \ & \ & \ \\ \ & \ & \ & \ &2& \ &-4& \ &4& \ &-4& \ &2\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &-4& \ &-2& \ &-2& \ &-4\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &3& \ &6& \ &3\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

有和 Schur 一样的取等。先扣外层的 \(a-b\) 因子 \(\mathrm{s(a2b2(a-b)4)+2s(a2b2(a2-b2)2)}\)，这样外圈就没东西了，可以降次：

\[\left[\begin{matrix}3& \ &-1& \ &-2& \ &-2& \ &-1& \ &3\ \\ \ &-1& \ &2& \ &1& \ &2& \ &-1\ & \ \\ \ & \ &-2& \ &1& \ &1& \ &-2\ & \ & \ \\ \ & \ & \ &-2& \ &2& \ &-2\ & \ & \ & \ \\ \ & \ & \ & \ &-1& \ &-1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &3\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

角上是个 \(3\)，所以拿掉三个 \(\mathrm{s(a3(a-b)(a-c))}\)。

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ &2& \ &-2& \ &-2& \ &2& \ & \textcolor{lightgray}{0}\ \\ \ &2& \ &-1& \ &1& \ &-1& \ &2\ & \ \\ \ & \ &-2& \ &1& \ &1& \ &-2\ & \ & \ \\ \ & \ & \ &-2& \ &-1& \ &-2\ & \ & \ & \ \\ \ & \ & \ & \ &2& \ &2\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

这个结构很像 \(\mathrm{s(a(a-b)(a-c))}\) 叠了三层（在三个维度上），所以拿掉一个 \(\mathrm{s(bc)s(a(a-b)(a-c))}\)（拿掉两个外圈就没东西了，内圈还是负的）：

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &1& \ &-1& \ &-1& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1\ & \ \\ \ & \ &-1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-1\ & \ & \ \\ \ & \ & \ &-1& \ & \textcolor{lightgray}{0}& \ &-1\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

你会惊奇的发现：内圈被拿光了！这下好办了：\(\mathrm{s(ab(a-b)(a2-b2))}\)。

也就是说这里又用到了一个结构 \(\mathrm{s(ab)s(a(a-b)(a-c))}\)，在五次和 Schur 有相同取等的不等式里面很有用，还是要背下来：

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &1& \ &-1& \ &-1& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &-1& \ &1& \ &-1& \ &1\ & \ \\ \ & \ &-1& \ &1& \ &1& \ &-1\ & \ & \ \\ \ & \ & \ &-1& \ &-1& \ &-1\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

4e.

\(\mathrm{s(a4+2ab3-2a3b+a2b2)}=\)

\[\left[\begin{matrix} 1& \ &-2& \ &1& \ &2& \ &1\ \\ \ &2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-2\ & \ \\ \ & \ &1& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ \\ \ & \ & \ &-2& \ &2\ & \ & \ & \ \\ \ & \ & \ & \ &1\ & \ & \ & \ & \ \end{matrix}\right] \]

非常简单的情况，留给读者。

4f.

\(\mathrm{s(a2)2-3s(a3b)}=\)

\[\left[\begin{matrix}1& \ &-3& \ &2& \ & \textcolor{lightgray}{0}& \ &1\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-3\ & \ \\ \ & \ &2& \ & \textcolor{lightgray}{0}& \ &2\ & \ & \ \\ \ & \ & \ &-3& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ &1\ & \ & \ & \ & \ \end{matrix}\right] \]

就是 Vasile。四次的话先求一下取等，然后算一下二次元。

a=0.54313396;
b=0.34929170;
c=0.10757434;
lindep([a^2-b^2,a*b-a*c,b*c-a*b], 7)

[-1, -1, -2]~

所以二次元是 \(\mathrm{a^2-b^2-ab-ac+2bc}\)。

发现原不等式就是 \(\mathrm{1/2s(a2-b2-ab-ac+2bc)2}\)，就做完了。

4g.

\(\mathrm{s(108a4+81a3b)-7s(a)4}=\)

\[\left[\begin{matrix}101& \ &53& \ &-42& \ &-28& \ &101\ \\ \ &-28& \ &-84& \ &-84& \ &53\ & \ \\ \ & \ &-42& \ &-84& \ &-42\ & \ & \ \\ \ & \ & \ &53& \ &-28\ & \ & \ & \ \\ \ & \ & \ & \ &101\ & \ & \ & \ & \ \end{matrix}\right] \]

非常非常松的不等式，取等条件很少。

看中间的 \(-84\) 不顺眼，分两份 \(-42\) 拿掉，也就是 \(\mathrm{42s(b(a-c)(a2-c2))}\)。

\[\left[\begin{matrix} 101& \ &11& \ &-42& \ &-70& \ &101\ \\ \ &-70& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &11\ & \ \\ \ & \ &-42& \ & \textcolor{lightgray}{0}& \ &-42\ & \ & \ \\ \ & \ & \ &11& \ &-70\ & \ & \ & \ \\ \ & \ & \ & \ &101\ & \ & \ & \ & \ \end{matrix}\right] \]

上边那个 \(-70\) 也很不顺眼，用最右边的 \(101\) 拿掉 \(35\)，也就是 \(\mathrm{35s(b2(a-b)2)}\)。

\[\left[\begin{matrix} 66& \ &11& \ &-77& \ & \textcolor{lightgray}{0}& \ &66\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &11\ & \ \\ \ & \ &-77& \ & \textcolor{lightgray}{0}& \ &-77\ & \ & \ \\ \ & \ & \ &11& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ &66\ & \ & \ & \ & \ \end{matrix}\right] \]

虽然是 Vasile 型的，但它没有特殊取等，所以可以随心配。扣掉 \(66\) 个 \(\mathrm{s(a)s(a(a-b)(a-c))}\)（这个结构也非常重要），得到

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ &11& \ &55& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &-66& \ &-66& \ &11\ & \ \\ \ & \ &55& \ &-66& \ &55\ & \ & \ \\ \ & \ & \ &11& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \end{matrix}\right] \]

要把它变对称，只需要扣掉 \(11\) 个这个结构：\(\mathrm{s(ac(b-c)2)}\)

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &-1& \ &-1& \ &1\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-1& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \end{matrix}\right] \]

拿掉之后就剩下中间的 \(55\) 包着 \(-55\)，用 \(\mathrm{s(a2(b-c)2)}\) 带过就好了。

5a.

\(\mathrm{2-s(a/(b+c))-p(a+b-c)/2/p(a)}=\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ &-2& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &-2& \ &1& \ &1& \ &-2& \ &1\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &-2& \ &1& \ &1& \ &-2\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &-2& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

先拿掉一个 \(\mathrm{s(ab)s(a2(a-b)(a-c))}\)。这个结构形状是一层 \(1\) 包着一层 \(-1\) 包着中间一个 \(3\)，刚好能帮我们把最外面的 \(1\) 扣掉。

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &-2& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &-1& \ &1& \ &1& \ &-1& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &1& \ &1& \ &-3& \ &1& \ &1\ & \ & \ \\ \ & \ & \ &-2& \ &1& \ &1& \ &-2\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &-1& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

这个结构很像 \(\mathrm{p((a-b)2)}\)，所以减掉一个。减掉之后就是 \(\mathrm{p(a)s(a(a-b)(a-c))}\)，做完了。

5b.

\(\mathrm{s(a/(b+c))+6s(ab)/s(a)2-7/2}=\)

\[\left[\begin{matrix} 2& \ &-1& \ &-1& \ &-1& \ &-1& \ &2\ \\ \ &-1& \ & \textcolor{lightgray}{0}& \ &2& \ & \textcolor{lightgray}{0}& \ &-1\ & \ \\ \ & \ &-1& \ &2& \ &2& \ &-1\ & \ & \ \\ \ & \ & \ &-1& \ & \textcolor{lightgray}{0}& \ &-1\ & \ & \ & \ \\ \ & \ & \ & \ &-1& \ &-1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &2\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

有 Schur 一样的取等。

尝试扣掉 \(\mathrm{s(a)s(a2(a-b)(a-c))}\)，发现只剩一个五次 Schur 了，结束。

8a.

\(\mathrm{s((a3+b3)/(c2+ab))-2s(a2/(b+c))}=\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &-1& \ & \textcolor{lightgray}{0}& \ &-1& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &3& \ &1& \ &-2& \ &-2& \ &1& \ &3& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &3& \ &-1& \ &1& \ & \textcolor{lightgray}{0}& \ &1& \ &-1& \ &3& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &1& \ &1& \ &1& \ &-5& \ &-5& \ &1& \ &1& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ &-1& \ &-2& \ & \textcolor{lightgray}{0}& \ &-5& \ & \textcolor{lightgray}{0}& \ &-2& \ &-1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &-2& \ &1& \ &1& \ &-2& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &-1& \ &1& \ &-1& \ &1& \ &-1\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ &1& \ &3& \ &3& \ &1\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

是的，十次，Schur 取等。去掉一个 \(\mathrm{s(a2b2)s(a(a-b)(a-c))2}\)，最外层就只剩 \([2,-4,2]\) 了，拿掉（\(\mathrm{2s(a3b3(a2-b2)2)}\)），再除以 \(\mathrm{p(a)}\)：

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ &4& \ &-6& \ &4& \ &4& \ &-6& \ &4& \ & \textcolor{lightgray}{0}\ \\ \ &4& \ &1& \ &4& \ &-13& \ &4& \ &1& \ &4\ & \ \\ \ & \ &-6& \ &4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &4& \ &-6\ & \ & \ \\ \ & \ & \ &4& \ &-13& \ & \textcolor{lightgray}{0}& \ &-13& \ &4\ & \ & \ & \ \\ \ & \ & \ & \ &4& \ &4& \ &4& \ &4\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &-6& \ &1& \ &-6\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &4& \ &4\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

外面的 \(-6\) 很不爽，直接 \(\mathrm{3s(ab(a3+b3)(a-b)2)}\)：

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ &1& \ &1& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &1& \ &4& \ &-13& \ &4& \ &1& \ &1\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &4& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &1& \ &-13& \ & \textcolor{lightgray}{0}& \ &-13& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &4& \ &4& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

中层的 \(1\) 全部拿去对付 \(-13\)（\(\mathrm{s(c3(a-b)(a3-b3))}\)）：

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &1& \ &4& \ &-11& \ &4& \ &1& \ &1\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &4& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &-11& \ & \textcolor{lightgray}{0}& \ &-11& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &4& \ &4& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

外层的 \(1\) 也拿去对付 \(-11\)（\(\mathrm{s(c(a3-b3)2)}\)），再除以 \(\mathrm{p(a)}\)：

\[\left[\begin{matrix} 1& \ &4& \ &-9& \ &4& \ &1\ \\ \ &4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &4\ & \ \\ \ & \ &-9& \ & \textcolor{lightgray}{0}& \ &-9\ & \ & \ \\ \ & \ & \ &4& \ &4\ & \ & \ & \ \\ \ & \ & \ & \ &1\ & \ & \ & \ & \ \end{matrix}\right] \]

拿掉一个 \(\mathrm{1/2s((a-b)4)}\) 之后发现刚刚好是 \(\mathrm{6s(ab(a-b)2)}\)，结束了。

8b.

\(\mathrm{s(a3)+3p(a)s(a2b)/s(ab2)-s(ab(a+b))}=\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &-1& \ &-1& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &-1& \ &1& \ &1& \ &-1& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &-1& \ &1& \ &-3& \ &1& \ &1\ & \ & \ \\ \ & \ & \ &-1& \ &1& \ &1& \ &-1\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &-1& \ &-1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

有 Schur 的取等。

和 5a 差不多的结构，一样先拿掉 \(\mathrm{p((a-b)2)}\) 和 \(\mathrm{p(a)s(a(a-b)(a-c))}\)，发现各拿一倍之后是 \(\mathrm{s(ab3(a-b)2)}\)，做完了。

9.

\(\mathrm{s(a3/(2a2+b2))-s(a)/3}=\)

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ &2& \ &-2& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-4& \ & \textcolor{lightgray}{0}& \ &-2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &1& \ &-2& \ &3& \ &3& \ &-4& \ &2\ & \ & \ \\ \ & \ & \ &-2& \ & \textcolor{lightgray}{0}& \ &3& \ & \textcolor{lightgray}{0}& \ &2\ & \ & \ & \ \\ \ & \ & \ & \ &2& \ &-4& \ &-2& \ &-2\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &2& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

不会配，先放在这。

10.

\(\mathrm{4s(a/(a+b))+(s(ab2)+abc)/(s(a2b)+abc)-7}\)

其实就是 \(\mathrm{p((a-b)2)}\)。这里不多赘述。

11.

\(\mathrm{s(a3+2a2b-3ab2)}=\)

\[\left[\begin{matrix} 1& \ &2& \ &-3& \ &1\ \\ \ &-3& \ & \textcolor{lightgray}{0}& \ &2\ & \ \\ \ & \ &2& \ &-3\ & \ & \ \\ \ & \ & \ &1\ & \ & \ & \ \end{matrix}\right] \]

一种非常天才的配方方法（作者没想到）：乘上 \(\mathrm{s((a-b)2)}\)：

\[\left[\begin{matrix}2& \ &2& \ &-8& \ &12& \ &-8& \ &2\ \\ \ &-8& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2\ & \ \\ \ & \ &12& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-8\ & \ & \ \\ \ & \ & \ &-8& \ & \textcolor{lightgray}{0}& \ &12\ & \ & \ & \ \\ \ & \ & \ & \ &2& \ &-8\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &2\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

这个 \([1,-4,6,-4,1]\) 非常明显，所以就是 \(\mathrm{2s(a(a-c)4)}\)。

12.

\(\mathrm{4s(a)3-27s(a2b)-27abc}=\)

\[\left[\begin{matrix}4& \ &-15& \ &12& \ &4\ \\ \ &12& \ &-3& \ &-15\ & \ \\ \ & \ &-15& \ &12\ & \ & \ \\ \ & \ & \ &4\ & \ & \ & \ \end{matrix}\right] \]

在 \((1,1,1)\) 以及 \((a,1,-a-1)\) 处取等，其中 \(a\) 是 \(a^3+2a^2-a-1=0\) 的根。

使用这里第八章第六节的定理，这里 \(p=-\frac{11}4,q=4,n=-\frac34\)，\(3(1+n)-(p^2+pq+q^2)=-\frac{189}{16}<0\)，使用第二条，求得 \(u=2,t=\frac34\)。

按照定理，我们可以扣掉一个 \(\mathrm{3s(ab(a+c-2b)2)}\)。

\[4\times\left[\begin{matrix}1& \ &- \frac{7}{2}& \ & \frac{9}{4}& \ &1& \ &1\ \\ \ &1& \ &- \frac{3}{4}& \ &- \frac{3}{4}& \ &- \frac{7}{2}\ & \ \\ \ & \ & \frac{9}{4}& \ &- \frac{3}{4}& \ & \frac{9}{4}\ & \ & \ \\ \ & \ & \ &- \frac{7}{2}& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ &1\ & \ & \ & \ & \ \end{matrix}\right] \]

再使用定理，\(p=-\frac72,q=1,n=\frac94,3(1+n)-(p^2+pq+q^2)=0\)，所以配方项里面第二项就没了，剩下 \(\mathrm{1/8s((2a^2-2b^2-(ac-ab)+4(bc-ab))2)}\)。

12.

做一个没那么强的：\(\mathrm{s(a3)/3-abc-22/15p(a-b)}\)。显然要升次，结果是

\[\left[\begin{matrix} 1& \ & \frac{27}{5}& \ & \textcolor{lightgray}{0}& \ &- \frac{17}{5}& \ &1\ \\ \ &- \frac{17}{5}& \ &-3& \ &-3& \ & \frac{27}{5}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-3& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \frac{27}{5}& \ &- \frac{17}{5}\ & \ & \ & \ \\ \ & \ & \ & \ &1\ & \ & \ & \ & \ \end{matrix}\right] \]

\(p=\frac{27}5,q=-\frac{17}5,n=0,3(1+n)-(p^2+pq+q^2)=-\frac{484}{25}<0\)，取近似 \(u=\frac{17}{39},t=4\)，这样还是可以配。

\[\left[\begin{matrix} 1& \ & \frac{7}{5}& \ & \frac{136}{39}& \ &- \frac{31637}{7605}& \ &1\ \\ \ &- \frac{31637}{7605}& \ &- \frac{2627}{1521}& \ &- \frac{2627}{1521}& \ & \frac{7}{5}\ & \ \\ \ & \ & \frac{136}{39}& \ &- \frac{2627}{1521}& \ & \frac{136}{39}\ & \ & \ \\ \ & \ & \ & \frac{7}{5}& \ &- \frac{31637}{7605}\ & \ & \ & \ \\ \ & \ & \ & \ &1\ & \ & \ & \ & \ \end{matrix}\right] \]

\(p=\frac75,q=-\frac{31637}{7605},n=\frac{136}{39},3(1+n)-(p^2+pq+q^2)=\frac{1142636}{57836025}>0\)，所以用第一种情况：

\[\mathrm{1/1041048450s((22815(a2-b2)-52627(ac-ab)+10343(bc-ab))2)}\\\;\\\kern{-150pt}\mathrm{+571318/173508075s(a2(b-c)2)} \]

14.

\(\mathrm{s(a2)2-4s(a)p(a-b)}=\)

\[\left[\begin{matrix} 1& \ &4& \ &2& \ &-4& \ &1\ \\ \ &-4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &4\ & \ \\ \ & \ &2& \ & \textcolor{lightgray}{0}& \ &2\ & \ & \ \\ \ & \ & \ &4& \ &-4\ & \ & \ & \ \\ \ & \ & \ & \ &1\ & \ & \ & \ & \ \end{matrix}\right] \]

\(p=4,q=-4,n=2,3(1+n)-(p^2+pq+q^2)=-7<0\)。\(u\) 有精确值 \(\sqrt2-1\)，这样 \(t=\sqrt2+1\)。

\[\left[\begin{matrix} 1& \ &3 - \sqrt{2}& \ &4& \ &-3 - \sqrt{2}& \ &1\ \\ \ &-3 - \sqrt{2}& \ &-2 + 2 \sqrt{2}& \ &-2 + 2 \sqrt{2}& \ &3 - \sqrt{2}\ & \ \\ \ & \ &4& \ &-2 + 2 \sqrt{2}& \ &4\ & \ & \ \\ \ & \ & \ &3 - \sqrt{2}& \ &-3 - \sqrt{2}\ & \ & \ & \ \\ \ & \ & \ & \ &1\ & \ & \ & \ & \ \end{matrix}\right] \]

\(p=3-\sqrt2,q=-3-\sqrt2,n=4,3(1+n)-(p^2+pq+q^2)=0\)。根据定理，这个东西等于 \(\mathrm{1/2s((a2-b2-(1+\sqrt2)(ac-ab)+(\sqrt2-1)(bc-ab))2)}\)。

15.

\(\mathrm{s(a/(b+c))+3abc/2/s(ab2)-2}\)
不会配，先放这。

16.

\(\mathrm{s((a2+bc)/(b2+c2))-5/2-4p(a)2/p(a2+b2)}\)

\[\left[\begin{matrix} 2& \ & \textcolor{lightgray}{0}& \ &-3& \ &2& \ &-3& \ & \textcolor{lightgray}{0}& \ &2\ \\ \ & \textcolor{lightgray}{0}& \ &2& \ &2& \ &2& \ &2& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &-3& \ &2& \ &-12& \ &2& \ &-3\ & \ & \ \\ \ & \ & \ &2& \ &2& \ &2& \ &2\ & \ & \ & \ \\ \ & \ & \ & \ &-3& \ &2& \ &-3\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &2\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

先随便扣 Schur，这里我扣了 \(\mathrm{2s(a4(a-b)(a-c))+s(ab)s(a2(a-b)(a-c))+p(a)s(a(a-b)(a-c))}\)。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &1& \ &-2& \ &2& \ &-2& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ & \textcolor{lightgray}{0}& \ &3& \ &3& \ & \textcolor{lightgray}{0}& \ &1\ & \ \\ \ & \ &-2& \ &3& \ &-18& \ &3& \ &-2\ & \ & \ \\ \ & \ & \ &2& \ &3& \ &3& \ &2\ & \ & \ & \ \\ \ & \ & \ & \ &-2& \ & \textcolor{lightgray}{0}& \ &-2\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

内圈用 \(\mathrm{3s(ab2c(a-c)2)}\)，外圈用 \(\mathrm{s(ab(a2+b2)(a-b)2)}\)，结束。

?

\(\mathrm{s(a/b)2-9s(a2)/s(ab)}=\)

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ &1& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &1& \ &-7& \ &2& \ &-7& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &1& \ &1& \ &2& \ &2& \ &2& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &2& \ &-7& \ &1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

外层 \(1\) 可以用 \(\mathrm{s(ab3(ab-c2)2)}\) 扣掉，这样就可以除以 \(p(a)\)。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &-5& \ &2& \ &-5& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &2& \ &2& \ &2\ & \ & \ \\ \ & \ & \ &2& \ &-5& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

再扣掉 \(\mathrm{s(a3(b-c)2)-1/2p(a)s((a-b)2)}\)：

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &-1& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &-4& \ &3& \ &-4& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &-1& \ &3& \ &3& \ &1\ & \ & \ \\ \ & \ & \ &1& \ &-4& \ &-1\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

如果要直接用配方解决这个，那么考虑把 \([1,-1,1]\) 拆成 \([1,-2,1]+[0,1,0]\)。那么配方式会是这样的：

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &-2& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{green}{-2}& \ & \textcolor{yellow}{4}& \ & \textcolor{blue}{-2}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &1& \ & \textcolor{blue}{-2}& \ & \textcolor{red}{1}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

也就是 \(\mathrm{a(a-b)2(b-c)2}\)。这样就做完了。

?

\(\mathrm{s(a(b+c)/(b2+c2))-2+8p(a)2/p(a2+b2)}=\)

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ &1& \ &-2& \ &2& \ &-2& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ & \textcolor{lightgray}{0}& \ &2& \ &2& \ & \textcolor{lightgray}{0}& \ &1\ & \ \\ \ & \ &-2& \ &2& \ &4& \ &2& \ &-2\ & \ & \ \\ \ & \ & \ &2& \ &2& \ &2& \ &2\ & \ & \ & \ \\ \ & \ & \ & \ &-2& \ & \textcolor{lightgray}{0}& \ &-2\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

外层是 \(\mathrm{s(ab(a2+b2)(a-b)2)}\)，减掉内层就是

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ &2& \ &2& \ & \textcolor{lightgray}{0}\ \\ \ &2& \ &-12& \ &2\ & \ \\ \ & \ &2& \ &2\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ \end{matrix}\right] \]

结束。

?

\(\mathrm{s(a/b)+8/3s(ab)/s(a2)-17/3}=\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &3& \ & \textcolor{lightgray}{0}& \ &3& \ & \textcolor{lightgray}{0}\ \\ \ &3& \ &-17& \ &11& \ &-17& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &11& \ &11& \ &3\ & \ & \ \\ \ & \ & \ &3& \ &-17& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &3\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

不会。先放在这里。

?

\(\mathrm{s(a2/b)-(6s(a2)-3s(ab))/s(a)}\)

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &-5& \ &3& \ &-5& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &1& \ &3& \ &3& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &-5& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

使用在喵喵喵 XV 提到的五次不对称配方技巧。乘上 \(\mathrm{s((a-b)2)}\) 之后就是 \(\mathrm{2s(a(b-c)2(ac+b2-abc)2)}\)。

?

\(\mathrm{s(a/b)+21s(ab)/s(a)2-10}\)

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &2& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &-8& \ &4& \ &-8& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &2& \ &4& \ &4& \ &1\ & \ & \ \\ \ & \ & \ &1& \ &-8& \ &2\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

和前面一样，\([0,0,1,2,1,0]\) 拆成 \([0,0,1,-2,1,0]\) 和 \([0,0,0,4,0,0]\) 就好了。

?

\(\mathrm{s(a(b+c)/(b2+bc+c2))-2-3p(a-b)2/p(a2+ab+b2)}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &1& \ &-4& \ &6& \ &-4& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &6& \ &-6& \ &-6& \ &6& \ &1\ & \ \\ \ & \ &-4& \ &-6& \ &18& \ &-6& \ &-4\ & \ & \ \\ \ & \ & \ &6& \ &-6& \ &-6& \ &6\ & \ & \ & \ \\ \ & \ & \ & \ &-4& \ &6& \ &-4\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

外层用 \(\mathrm{s(ab(a-b)4)}\) 扣掉，内层是 \(\mathrm{6p(a)s(a(a-b)(a-c))}\)。

?

\(\mathrm{s((a2+kbc)/(b2-bc+c2))-2-k-(1+2k)p(a)s(a3)/3/p(a2-ab+b2)}\)

这里我们只证 \(k=0,1\)。

\(k=0\)

\[\left[\begin{matrix}3& \ &-3& \ &-3& \ &6& \ &-3& \ &-3& \ &3\ \\ \ &-3& \ &8& \ &-3& \ &-3& \ &8& \ &-3\ & \ \\ \ & \ &-3& \ &-3& \ &3& \ &-3& \ &-3\ & \ & \ \\ \ & \ & \ &6& \ &-3& \ &-3& \ &6\ & \ & \ & \ \\ \ & \ & \ & \ &-3& \ &8& \ &-3\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &-3& \ &-3\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &3\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

Schur 取等。

先扣掉三个 \(\mathrm{s(a(a-b)(a-c))2}\)。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &3& \ & \textcolor{lightgray}{0}& \ &-6& \ & \textcolor{lightgray}{0}& \ &3& \ & \textcolor{lightgray}{0}\ \\ \ &3& \ &-16& \ &15& \ &15& \ &-16& \ &3\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &15& \ &-42& \ &15& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &-6& \ &15& \ &15& \ &-6\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &-16& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &3& \ &3\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

然后扣掉另一个重要的结构 \(\mathrm{s(a(b+c)(a-b)2(a-c)2)}\)。这个结构等于 \(\mathrm{s(ab(a2+b2)(a-b)2)-4p(a)s(a(a-b)(a-c))}\)，有 Schur 取等。

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ &1& \ &4& \ &-10& \ &4& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &-8& \ &7& \ &7& \ &-8& \ &1\ & \ \\ \ & \ &4& \ &7& \ &-18& \ &7& \ &4\ & \ & \ \\ \ & \ & \ &-10& \ &7& \ &7& \ &-10\ & \ & \ & \ \\ \ & \ & \ & \ &4& \ &-8& \ &4\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

这样我们就可以扣四次 Schur 了。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &5& \ &-10& \ &5& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &-7& \ &7& \ &7& \ &-7& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &5& \ &7& \ &-21& \ &7& \ &5\ & \ & \ \\ \ & \ & \ &-10& \ &7& \ &7& \ &-10\ & \ & \ & \ \\ \ & \ & \ & \ &5& \ &-7& \ &5\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

很像 \(\mathrm{p(a-b)2}\) 的结构，所以扣掉。剩下的就是 \(\mathrm{3p(a)s(a(a-b)(a-c))}\)。

\(k=1\)

\[\left[\begin{matrix}1& \ &-1& \ &-2& \ &4& \ &-2& \ &-1& \ &1\ \\ \ &-1& \ &4& \ &-2& \ &-2& \ &4& \ &-1\ & \ \\ \ & \ &-2& \ &-2& \ &3& \ &-2& \ &-2\ & \ & \ \\ \ & \ & \ &4& \ &-2& \ &-2& \ &4\ & \ & \ & \ \\ \ & \ & \ & \ &-2& \ &4& \ &-2\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &-1& \ &-1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &1\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

先扣掉一个 \(\mathrm{s(a(a-b)(a-c))2}\)，然后最外圈就只剩两个对称的 \(1\) 了，用 \(\mathrm{s(ab(a-b)4)}\) 这个最强的结构扣掉。

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &3& \ &-6& \ &3& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &-4& \ &4& \ &4& \ &-4& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &3& \ &4& \ &-12& \ &4& \ &3\ & \ & \ \\ \ & \ & \ &-6& \ &4& \ &4& \ &-6\ & \ & \ & \ \\ \ & \ & \ & \ &3& \ &-4& \ &3\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

热力图上很像 \(\mathrm{p(a-b)2}\)，所以拿掉三个。剩下的就是两个 \(\mathrm{p(a)s(a(a-b)(a-c))}\)。

Ukraine 2006

\(\mathrm{3s(a3)+3abc-4s(a2b)}\)

\[\left[\begin{matrix}3& \ &-4& \ & \textcolor{lightgray}{0}& \ &3\ \\ \ & \textcolor{lightgray}{0}& \ &3& \ &-4\ & \ \\ \ & \ &-4& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &3\ & \ & \ & \ \end{matrix}\right] \]

拿掉两个 \(\mathrm{s(a(a-b)(a-c))}\)，不能继续配，升次。其实只扣一个的话还有另一种方法。

\[\left[\begin{matrix} 1& \ &-1& \ & \textcolor{lightgray}{0}& \ &3& \ &1\ \\ \ &3& \ &-3& \ &-3& \ &-1\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-3& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &-1& \ &3\ & \ & \ & \ \\ \ & \ & \ & \ &1\ & \ & \ & \ & \ \end{matrix}\right] \]

扣掉一个 \(\mathrm{s(a2(a-b)(a-c))}\) 之后就是 \(\mathrm{4s(ac(a-b)2)}\)。

19.

\(\mathrm{s(a)s(1/a)+ks(ab)/s(a2)-9-k},k=4\sqrt2\)

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ &1& \ &1& \ &1& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &-6 - 4 \sqrt{2}& \ &2 + 4 \sqrt{2}& \ &-6 - 4 \sqrt{2}& \ &1\ & \ \\ \ & \ &1& \ &2 + 4 \sqrt{2}& \ &2 + 4 \sqrt{2}& \ &1\ & \ & \ \\ \ & \ & \ &1& \ &-6 - 4 \sqrt{2}& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

不会，但它有配方 \(\mathrm{s(a(b-c)2((-1-\sqrt2)a+b+c)2)}\)。

20.

\(\mathrm{3/5-s(bc/(b2+c2+3a2))}\)

\[\left[\begin{matrix}9& \ &-15& \ &39& \ &-50& \ &39& \ &-15& \ &9\ \\ \ &-15& \ &-5& \ &-20& \ &-20& \ &-5& \ &-15\ & \ \\ \ & \ &39& \ &-20& \ &114& \ &-20& \ &39\ & \ & \ \\ \ & \ & \ &-50& \ &-20& \ &-20& \ &-50\ & \ & \ & \ \\ \ & \ & \ & \ &39& \ &-5& \ &39\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &-15& \ &-15\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &9\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

定向配方可以扣出来 \(\mathrm{s((a2+4bc)(a-b)(a-c)(3a-2b)(3a-2c))}\)，因为 \(f(a,1,1)=(a^2+4)(3a-2)^2(a-1)^2\)。扣掉之后只剩 \(\mathrm{33p(a-b)2}\) 了。

22.

\(\mathrm{4s(a)6-27(s(a2b)+2s(ab2))2}\)

暂时不会配。

23.

\(\mathrm{4s(a)4-27s(a3b)-27s(a2b2)-27p(a)s(a)}\)
我们证明一个更强的：再扣掉 \(\mathrm{27p(a)s(a)}\)。

\[\left[\begin{matrix}1& \ &- \frac{11}{4}& \ &- \frac{3}{4}& \ &4& \ &1\ \\ \ &4& \ &- \frac{3}{2}& \ &- \frac{3}{2}& \ &- \frac{11}{4}\ & \ \\ \ & \ &- \frac{3}{4}& \ &- \frac{3}{2}& \ &- \frac{3}{4}\ & \ & \ \\ \ & \ & \ &- \frac{11}{4}& \ &4\ & \ & \ & \ \\ \ & \ & \ & \ &1\ & \ & \ & \ & \ \end{matrix}\right] \]

这就和 #12 一样了。

24.

我们证明 \(k=\frac{11}4\) 的情况。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &4& \ & \textcolor{lightgray}{0}& \ &4& \ & \textcolor{lightgray}{0}\ \\ \ &4& \ &-23& \ &15& \ &-23& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &15& \ &15& \ &4\ & \ & \ \\ \ & \ & \ &4& \ &-23& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &4\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

先放着。

25.

我们证明 \(k=\frac{19}{12}\) 的情况。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &6& \ &6& \ & \textcolor{lightgray}{0}\ \\ \ &6& \ &-31& \ &19& \ &-31& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &6& \ &19& \ &19& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &-31& \ &6\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &6\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

APMO 2004

\(\mathrm{p((a/b)2+2)-9s(a/c)}=\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ & \textcolor{lightgray}{0}& \ &4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-9& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &4& \ & \textcolor{lightgray}{0}& \ &9& \ &-9& \ &2\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &-9& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ &2& \ & \textcolor{lightgray}{0}& \ &4\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

通过扣 \(\mathrm{s(a2(b2-ac)2)}\) 和 \(\mathrm{s(a2b2(a-c)2)}\) 实现 \(a^4b^2,a^2b^4\) 的系数都是 \(1\)。

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &1& \ & \textcolor{lightgray}{0}& \ &6& \ &-4& \ &1\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &-4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

然后扣 \(\mathrm{p(a-b)2}\)。

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &2& \ &-6& \ &-2& \ &2& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-2& \ &12& \ &-6& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &2& \ &-6& \ &-2& \ &2\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &2& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

结构很像六芒星，所以扣掉 \(\mathrm{2s(ab(a-c)2(b-c)2)}\)。剩下是 \(\mathrm{4p(a)}\) 倍的

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &-3& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ \end{matrix}\right] \]

升次套一下定理就做完了。

26.

\(\mathrm{s(a/(b3+c3))-18/(5s(a2)-s(ab))}=\)

\[\left[\begin{matrix}5& \ &-1& \ &5& \ &-13& \ &4& \ &4& \ &-13& \ &5& \ &-1& \ &5\ \\ \ &-1& \ &-1& \ & \textcolor{lightgray}{0}& \ &4& \ &-3& \ &4& \ & \textcolor{lightgray}{0}& \ &-1& \ &-1\ & \ \\ \ & \ &5& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &4& \ &4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &5\ & \ & \ \\ \ & \ & \ &-13& \ &4& \ &4& \ &-21& \ &4& \ &4& \ &-13\ & \ & \ & \ \\ \ & \ & \ & \ &4& \ &-3& \ &4& \ &4& \ &-3& \ &4\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &4& \ &4& \ & \textcolor{lightgray}{0}& \ &4& \ &4\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &-13& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-13\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ &5& \ &-1& \ &5\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ &-1& \ &-1\ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ &5\ & \ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

非常吓人的几乎满系数的 \(9\) 次不等式。暂时不配。

28.

\(\mathrm{27(s(a3)+7abc)2-100s(ab)3}\)

先扣掉 \(\mathrm{27s(a4(a-b)(a-c))}\)。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &27& \ & \textcolor{lightgray}{0}& \ &-46& \ & \textcolor{lightgray}{0}& \ &27& \ & \textcolor{lightgray}{0}\ \\ \ &27& \ &351& \ &-300& \ &-300& \ &351& \ &27\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-300& \ &723& \ &-300& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &-46& \ &-300& \ &-300& \ &-46\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &351& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &27& \ &27\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

中间的 \(-46\) 很不爽，扣 \(23\) 个 \(\mathrm{s(ab(a2-b2)2)}\)。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &4& \ & \textcolor{lightgray}{0}\ \\ \ &4& \ &351& \ &-300& \ &-300& \ &351& \ &4\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-300& \ &723& \ &-300& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &-300& \ &-300& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &351& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &4& \ &4\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

两边的 \(4\) 扣掉。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &347& \ &-288& \ &-300& \ &347& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-300& \ &723& \ &-288& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &-288& \ &-300& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &347& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

然后因为这个不等式实在太松，可以直接扣掉 \(347\) 个 \(\mathrm{p(a)s(a(a-b)(a-c))}\)。剩下的是 \(\mathrm{p(a)}\) 倍的

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ &59& \ &47& \ & \textcolor{lightgray}{0}\ \\ \ &47& \ &-318& \ &59\ & \ \\ \ & \ &59& \ &47\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ \end{matrix}\right] \]

再扣 \(\mathrm{47s(a(b-c)2)}\) 就是简单的 \(\mathrm{s(a2b-abc)}\)，定理配一下就好。

31.

\(\mathrm{s((a/(b+c))2)+10p(a)/p(a+b)-2}\)

\[\left[\begin{matrix}1& \ &2& \ &-1& \ &-4& \ &-1& \ &2& \ &1\ \\ \ &2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2\ & \ \\ \ & \ &-1& \ & \textcolor{lightgray}{0}& \ &3& \ & \textcolor{lightgray}{0}& \ &-1\ & \ & \ \\ \ & \ & \ &-4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-4\ & \ & \ & \ \\ \ & \ & \ & \ &-1& \ & \textcolor{lightgray}{0}& \ &-1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &2& \ &2\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &1\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

减掉 Schur 的部分 \(\mathrm{s(a4(a-b)(a-c))+s(ab)s(a2(a-b)(a-c))}\) 就剩边上的平方了。

文献 2

2.

\(\mathrm{6s(a3)+s(a)3-5s(a2)s(a)}\)

\[\left[\begin{matrix}2& \ &-2& \ &-2& \ &2\ \\ \ &-2& \ &6& \ &-2\ & \ \\ \ & \ &-2& \ &-2\ & \ & \ \\ \ & \ & \ &2\ & \ & \ & \ \end{matrix}\right] \]

其实就是两个 Schur。这里讲一下三次 Schur 怎么配方。

显然必须要升次。

\[\left[\begin{matrix}1& \ & \textcolor{lightgray}{0}& \ &-2& \ & \textcolor{lightgray}{0}& \ &1\ \\ \ & \textcolor{lightgray}{0}& \ &1& \ &1& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &-2& \ &1& \ &-2\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ &1\ & \ & \ & \ & \ \end{matrix}\right] \]

可以用定理，也可以用定向配方。

用定理的话 \(p=q=0,n=-2\)，所以先拿掉第二种。显然 \(u=t=1\)，所以拿掉 \(\mathrm{s(ab(a-b)2)}\)。剩下的是 \(\mathrm{s((a2-b2+bc-ac)2)}\)。

这个东西的四个根可以被 \(a-b=0\) 和 \(a+b-c=0\) 覆盖，所以扣掉 \(\mathrm{1/2s((a-b)2(a+b-c)2)}\)（其实这个就是四次 Schur）。

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ &1& \ &-2& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1\ & \ \\ \ & \ &-2& \ & \textcolor{lightgray}{0}& \ &-2\ & \ & \ \\ \ & \ & \ &1& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \end{matrix}\right] \]

就做完了。

6.

先做一下齐次换元。

\(\mathrm{s((a/b)3/(1+b/c)/(1+c/a))-3/4}\)

确实是十二次。最外层的 \(4\) 扣掉之后剩下 \(\mathrm{p(a)}\) 倍的

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &12& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-3& \ &-3& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &12& \ &-4& \ &-3& \ &-6& \ &-3& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-3& \ &-3& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-4& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &12\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

中层负项处理掉（\(\mathrm{2s(a4c3(a-b)2)}\)），外层用 \([1,-1,-1,1]\) 导到内层（\(\mathrm{5s(b3(ab+c2)(ab-c2)2)}\)）。剩下的是 \(\mathrm{p(a)}\) 倍的

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &5& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &5& \ &-3& \ &-5& \ &5& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-5& \ &-6& \ &-3& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &5& \ &-3& \ &-5& \ &5\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &5& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

可以升次然后用伴随二次元，但这里这个结构很松，我们先减掉 \(\mathrm{5s(ab(a-c)2(b-c)2)}\)（也就是六芒星结构），剩下的是 \(\mathrm{2p(a)}\) 倍的

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &6& \ &5& \ & \textcolor{lightgray}{0}\ \\ \ &5& \ &-33& \ &6\ & \ \\ \ & \ &6& \ &5\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ \end{matrix}\right] \]

拿掉 \(\mathrm{5s(c(a-b)2)}\) 之后就是很经典的 \(\mathrm{s(a2b-abc)}\) 了。

7.

做一下齐次化。

\(\mathrm{1-s(1/(1+a/b+b/c))}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ &-2& \ &1& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &1& \ & \textcolor{lightgray}{0}& \ &-2& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

拿掉一个六芒星，剩下的是 \(\mathrm{p(a)}\) 倍的

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &3& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &-12& \ &3\ & \ \\ \ & \ &3& \ &1\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ \end{matrix}\right] \]

和上面做法就一样了。

8.

其实就是 \(\mathrm{s(a2b-abc)}\)。

9.

\(\mathrm{s(1/a/(b+c))-27/2/s(a)2}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ &4& \ &2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &6& \ &-5& \ &-5& \ &6& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &2& \ &-5& \ &-12& \ &-5& \ &2\ & \ & \ \\ \ & \ & \ &4& \ &-5& \ &-5& \ &4\ & \ & \ & \ \\ \ & \ & \ & \ &2& \ &6& \ &2\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

先拿掉两个 \(\mathrm{2p(a-b)2}\)。

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &8& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &10& \ &-9& \ &-9& \ &10& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-9& \ & \textcolor{lightgray}{0}& \ &-9& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &8& \ &-9& \ &-9& \ &8\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &10& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

按照惯例，拿掉八个六芒星（\(\mathrm{s(ab(a-c)2(b-c)2)}\)）。剩下的是 \(\mathrm{p(a)}\) 倍的

\[\left[\begin{matrix}2& \ &15& \ &15& \ &2\ \\ \ &15& \ &-96& \ &15\ & \ \\ \ & \ &15& \ &15\ & \ & \ \\ \ & \ & \ &2\ & \ & \ & \ \end{matrix}\right] \]

拿掉两个 Schur 之后就特别简单了。

文献 3

1.1.5

\(\mathrm{s(a2)3-s(a)3p(a)}\)

\[\left[\begin{matrix}1& \ & \textcolor{lightgray}{0}& \ &3& \ & \textcolor{lightgray}{0}& \ &3& \ & \textcolor{lightgray}{0}& \ &1\ \\ \ & \textcolor{lightgray}{0}& \ &-1& \ &-3& \ &-3& \ &-1& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &3& \ &-3& \ & \textcolor{lightgray}{0}& \ &-3& \ &3\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &-3& \ &-3& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ &3& \ &-1& \ &3\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &1\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

扣掉 Schur 和 \(\mathrm{p(a-b)2}\) 结构，变成六芒星结构。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &8& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &7& \ &-11& \ &-11& \ &7& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-11& \ &21& \ &-11& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &8& \ &-11& \ &-11& \ &8\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &7& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

扣掉 \(7\) 个六芒星和一个 Schur。剩下的是 \(\mathrm{11p(a)}\) 倍的

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &1& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &-6& \ &1\ & \ \\ \ & \ &1& \ &1\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ \end{matrix}\right] \]

1.3.1

\(\mathrm{s(a/b)3p(a)-s(a)3}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &3& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ &-3& \ &-3& \ &2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &1& \ & \textcolor{lightgray}{0}& \ &-3& \ & \textcolor{lightgray}{0}& \ &-3& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &3& \ &-3& \ &-3& \ &3& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

扣掉一个斜过来的 Schur（\(\mathrm{s(ab2(ab2-bc2)(ab2-ca2))}\)）。剩下的是 \(\mathrm{p(a)}\) 倍的

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &3& \ &-3& \ &-3& \ &3& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-3& \ &-3& \ &-3& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &4& \ &-3& \ &-3& \ &4\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &3& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

按照惯例扣掉 \(3\) 个六芒星和一个 Schur，剩下 \(\mathrm{7p(a)}\) 倍的 \(\mathrm{s(c(a-b)2)}\)，结束。

1.3.3.

上界

\(\mathrm{3s(a3)-s(a)s(a2)}\)

\[\left[\begin{matrix}2& \ &-1& \ &-1& \ &2\ \\ \ &-1& \ & \textcolor{lightgray}{0}& \ &-1\ & \ \\ \ & \ &-1& \ &-1\ & \ & \ \\ \ & \ & \ &2\ & \ & \ & \ \end{matrix}\right] \]

其实就是 \(\mathrm{s((a+b)(a-b)2)}\)。

下界

更简单，读者自己配。

1.3.5

证明一下 \(n=0\) 和 \(n=3\) 的情况。

\(n=0\)

\(\mathrm{s(a2/b)-3s(ab)/s(a)}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &1& \ &-3& \ &1& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &1& \ &-3& \ &-3& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &1& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

先拿掉半个 \(\mathrm{p(a)s((a-b)2)}\)。剩下的按照三种颜色构成三个 \([1,-2,1]\)：

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &\textcolor{blue}{1}& \ &\textcolor{green}{1}& \ & \textcolor{lightgray}{0}\ \\ \ &\textcolor{red}{1}& \ & \textcolor{lightgray}{0}& \ &\textcolor{green}{-2}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &\textcolor{green}{1}& \ &\textcolor{red}{-2}& \ &\textcolor{blue}{-2}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &\textcolor{red}{1}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &\textcolor{blue}{1}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

也就是 \(\mathrm{s(c(a2-bc)2)}\)。

\(n=3\)

\(\mathrm{s(a2/b)+9s(ab)3/s(a2)/s(a)3-6s(ab)/s(a)}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &3& \ &4& \ &4& \ &3& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &3& \ &-2& \ &-5& \ &6& \ &-5& \ & \textcolor{lightgray}{0}& \ &3& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &3& \ & \textcolor{lightgray}{0}& \ &-24& \ &1& \ & \textcolor{lightgray}{0}& \ &-24& \ &-2& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &4& \ &-5& \ & \textcolor{lightgray}{0}& \ &30& \ &1& \ &-5& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ &4& \ &6& \ &1& \ & \textcolor{lightgray}{0}& \ &6& \ &3\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &3& \ &-5& \ &-24& \ &-5& \ &4\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &1& \ &-2& \ & \textcolor{lightgray}{0}& \ &4\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &3& \ &3\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

边缘上的东西一阵乱扣，最终扣掉了 \(\mathrm{3s(b5(a-c)4)+s(a2c(a3-b2c)2)+2s(b3(ab+c2)(ab-c2)2)+s(a4b3(a-b)2)+2s(a5(ac-b2)2)+s(a3c2(a2-b2)2)}\)。剩下来的是 \(\mathrm{p(a)}\) 倍的

\[\left[\begin{matrix}3& \ &2& \ &7& \ &8& \ &7& \ & \textcolor{lightgray}{0}& \ &3\ \\ \ & \textcolor{lightgray}{0}& \ &-36& \ & \textcolor{lightgray}{0}& \ &-1& \ &-36& \ &2\ & \ \\ \ & \ &7& \ &-1& \ &30& \ & \textcolor{lightgray}{0}& \ &7\ & \ & \ \\ \ & \ & \ &8& \ & \textcolor{lightgray}{0}& \ &-1& \ &8\ & \ & \ & \ \\ \ & \ & \ & \ &7& \ &-36& \ &7\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &2& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &3\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

后面大多都针对中间的负项。\(\mathrm{7s(a4(b-c)2)+2s(a3b(a-c)2)}\)

\[\left[\begin{matrix}3& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &8& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &3\ \\ \ & \textcolor{lightgray}{0}& \ &-18& \ & \textcolor{lightgray}{0}& \ &-3& \ &-18& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-3& \ &30& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &8& \ & \textcolor{lightgray}{0}& \ &-3& \ &8\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &-18& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &3\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

然后发现扣不了线性的东西了，所以扣 \(\mathrm{3s(a(a-b)(a-c))2}\)。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &5& \ &5& \ &-6& \ &5& \ &5& \ & \textcolor{lightgray}{0}\ \\ \ &5& \ &-38& \ &14& \ &11& \ &-38& \ &5\ & \ \\ \ & \ &5& \ &11& \ &-3& \ &14& \ &5\ & \ & \ \\ \ & \ & \ &-6& \ &14& \ &11& \ &-6\ & \ & \ & \ \\ \ & \ & \ & \ &5& \ &-38& \ &5\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &5& \ &5\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

内层很像负的 Schur，所以拿掉上文提到的 \(\mathrm{s(a(b+c)(a-b)2(a-c)2)}\)。

然后扣掉另一个重要的结构 \(\mathrm{s(a(b+c)(a-b)2(a-c)2)}\)。这个结构等于 \(\mathrm{s(ab(a2+b2)(a-b)2)-4p(a)s(a(a-b)(a-c))}\)，有 Schur 取等。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &15& \ &-16& \ &15& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &-18& \ &-6& \ &-9& \ &-18& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &15& \ &-9& \ &57& \ &-6& \ &15\ & \ & \ \\ \ & \ & \ &-16& \ &-6& \ &-9& \ &-16\ & \ & \ & \ \\ \ & \ & \ & \ &15& \ &-18& \ &15\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

然后就是经典的拿 \(\mathrm{p(a-b)2}\) 和六芒星的过程了。拿完之后长这个样：

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-3& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-3& \ &3& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &2& \ & \textcolor{lightgray}{0}& \ &-3& \ &2\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

拿掉一个 Schur 之后就是 \([1,-2,1]\) 了。

2.1.5

\(\mathrm{s((a/(a+2b))2)-1/3}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &5& \ &8& \ &8& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &8& \ &-11& \ &-8& \ &8& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &8& \ &-8& \ &-30& \ &-11& \ &5\ & \ & \ \\ \ & \ & \ &8& \ &-11& \ &-8& \ &8\ & \ & \ & \ \\ \ & \ & \ & \ &5& \ &8& \ &8\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

减掉对应的 \(\mathrm{p(a-b)2}\) 和六芒星之后剩下的长这样：

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &11& \ &12& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &1& \ &12& \ &-72& \ &11& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &11& \ &12& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

用 \(\mathrm{s(ac2(a+b)(a-b)2)}\) 拿掉 \(1\)。剩下 \(\mathrm{p(a)}\) 倍的

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &10& \ &13& \ & \textcolor{lightgray}{0}\ \\ \ &13& \ &-69& \ &10\ & \ \\ \ & \ &10& \ &13\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ \end{matrix}\right] \]

这里不再赘述。

2.1.6.

\(\mathrm{s(1/b/(a+b))3-27/8/p(a)2}\)

十二次，这里不展示。

拿掉外圈项（\(\mathrm{8s(a6(ab+c2)(ab-c2)2)+15s(ab5(a2-c2)2(a2+c2))+6s(ab5(a3-bc2)2)+24s(b2c2(a4-b3c)2)+3s(a3b3(a2b-c3)2)}\)）之后剩下的是 \(\mathrm{p(a)}\) 倍的

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &48& \ &111& \ &87& \ & \textcolor{lightgray}{0}& \ &5& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &8& \ &111& \ &108& \ &9& \ & \textcolor{lightgray}{0}& \ &39& \ &8& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &39& \ &7& \ &-192& \ &-189& \ &7& \ &111& \ &48\ & \ & \ \\ \ & \ & \ &5& \ & \textcolor{lightgray}{0}& \ &-189& \ &-456& \ &-192& \ &108& \ &111\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &9& \ &-192& \ &-189& \ &9& \ &87\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &87& \ &108& \ &7& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &111& \ &111& \ &39& \ &5\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ &48& \ &8& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

扣掉其它所有负项之后，留下来了这个：

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &82& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &8& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &38& \ &8& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &38& \ &7& \ &5& \ &12& \ &7& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &12& \ &-456& \ &5& \ & \textcolor{lightgray}{0}& \ &82\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &5& \ &12& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &7& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &82& \ & \textcolor{lightgray}{0}& \ &38& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &8& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

然后扣掉最外层的 \(82\)（\(\mathrm{82s(a2b3(a-c)4)}\)）得到 \(\mathrm{p(a)}\) 倍的

\[\left[\begin{matrix}8& \ & \textcolor{lightgray}{0}& \ &328& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &38& \ &8\ \\ \ &38& \ &7& \ &-487& \ &-70& \ &7& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-70& \ &528& \ &-487& \ &328\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &-487& \ &-70& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ &328& \ &7& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &38\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &8\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

拿掉 \(\mathrm{8s((a3-b2c)2)+35s(b2(a2-bc)2)+7s(ac(a2-b2)2)+457/2s(a2b2(a-c)2)}\)。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &43& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &62& \ & \textcolor{lightgray}{0}\ \\ \ &62& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-315& \ & \textcolor{lightgray}{0}& \ &43\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ &43& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &62\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

再拿 \(\mathrm{43p(a-b)2+43/2s(ab3(a-b)2)+57/2s(ab(ab-c2)2)+43/2s(ab(b-c)4)+36s(ab(a-c)2(b-c)2)+86p(a)s(a(a-b)(a-c))+19s(ac(a-b)4)}\)，剩下的是 \(\mathrm{3p(a)}\) 倍的

\[\left[\begin{matrix}19& \ &-45& \ &90& \ &19\ \\ \ &90& \ &-192& \ &-45\ & \ \\ \ & \ &-45& \ &90\ & \ & \ \\ \ & \ & \ &19\ & \ & \ & \ \end{matrix}\right] \]

至于为什么这样拿，可以自己模拟一下。剩下的这个东西拿 Schur 会直接原地趋势，所以升次套定理。

2.2.4

\(\mathrm{s(a6/(b2+c2))-p(a)s(a)/2}\)
十次。用 Schur 和线性项拿走外圈项（\(\mathrm{2s(a6(a2-b2)(a2-c2))+4s(a2b2)s(a2(a2-b2)(a2-c2))+1/4s(a6(b-c)4)+15/4s(b4(a-c)6)}\)），降次。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &43& \ &43& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &5& \ &-2& \ &-233& \ &-2& \ &5& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-2& \ &146& \ &146& \ &-2& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &43& \ &-233& \ &146& \ &-233& \ &43\ & \ & \ & \ \\ \ & \ & \ & \ &43& \ &-2& \ &-2& \ &43\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &5& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

扣掉六芒星（\(\mathrm{s(a)s(ab(a-c)2(b-c)2)}\)），再扣掉 \(\mathrm{42s(a3(b-c)4)}\)。剩下的是 \(\mathrm{4p(a)}\) 倍的

\[\left[\begin{matrix} 1& \ & \textcolor{lightgray}{0}& \ &27& \ & \textcolor{lightgray}{0}& \ &1\ \\ \ & \textcolor{lightgray}{0}& \ &-28& \ &-28& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &27& \ &-28& \ &27\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ &1\ & \ & \ & \ & \ \end{matrix}\right] \]

扣掉三次 Schur 就剩中间的 \(29\) 个 \([1/2,-1,1/2]\) 了。

3.1.7

\(\mathrm{s(a2/(b2+bc))s(a/(a+c))-(s(a/(b+c)))2}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &3& \ &2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &-1& \ &-1& \ &5& \ &4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &2& \ &4& \ &-1& \ &-5& \ &-4& \ &-1& \ &1& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &3& \ &5& \ &-4& \ &-12& \ &-5& \ &-1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &-1& \ &-5& \ &-4& \ &-1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &-1& \ &-1& \ &5& \ &1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &1& \ &4& \ &3\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

用 \(\mathrm{s(ab2(ab+c2)(ab-c2)2)+s(b3(ab+c2)(ab-c2)2)+1/2s(b2(a+b)(ab2-c3)2)+s(a2b(b3-c3)2)+1/2s(a3c2(a-b)4)}\) 扣掉外圈项，剩下

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &10& \ &9& \ & \textcolor{lightgray}{0}\ \\ \ &9& \ &-6& \ & \textcolor{lightgray}{0}& \ &-5& \ &-6& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &10& \ &-5& \ &-24& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-5& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &-6& \ &10\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &9\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

通过 \(\mathrm{2s(ac(a-b)4)+5s(b2(ab-c2)2)+s(ac(a2-b2)2)+6s(a3c(a-b)2)}\) 拿掉外圈项。

\[\left[\begin{matrix}11& \ &-6& \ &3& \ &11\ \\ \ &3& \ &-24& \ &-6\ & \ \\ \ & \ &-6& \ &3\ & \ & \ \\ \ & \ & \ &11\ & \ & \ & \ \end{matrix}\right] \]

拿掉 Schur 之后就是很简单的情况了。

3.1.3

\(\mathrm{s(a8)/p(a3)-s(1/a)}\)
拿掉半个 \(\mathrm{s((a4-b4)2)+s(a4(b2-c2)2)}\) 之后就没了。

3.1.7

\(\mathrm{s(ab2)3-s(a)3p(a)2}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &3& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ &-3& \ &-3& \ &2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &1& \ & \textcolor{lightgray}{0}& \ &-3& \ & \textcolor{lightgray}{0}& \ &-3& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &3& \ &-3& \ &-3& \ &3& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

中间拿掉 Schur 和六芒星之后是

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &4& \ &4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &1& \ & \textcolor{lightgray}{0}& \ &4& \ &-27& \ &4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &4& \ &4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

通过 \(\mathrm{s(a2c3(a2-b2)2)}\) 把外面的 \(1\) 去掉之后就没什么了。

3.2.7

只证明上界。下界次数太高。

\(\mathrm{3s(ab/c2/(a+b))-s(1/c2/(a+b))s(ab)}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-2& \ &-2& \ &-2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &1& \ &-2& \ &-2& \ &1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ &2& \ &1& \ &-2& \ &1& \ &2\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

先旋转一下。

\[\left[\begin{matrix} 2& \ &1& \ &-2& \ &1& \ &2\ \\ \ &1& \ &-2& \ &-2& \ &1\ & \ \\ \ & \ &-2& \ &-2& \ &-2\ & \ & \ \\ \ & \ & \ &1& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ &2\ & \ & \ & \ & \ \end{matrix}\right] \]

刚好是 \(\mathrm{2s(a)s(a(a-b)(a-c))+s(ab(a-b)2)+2s(a2(b-c)2)}\)。

3.3.4

\(\mathrm{s(a2)/s(ab)-2/3s(a/(b+c))}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &1& \ &1& \ &1& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ & \textcolor{lightgray}{0}& \ &-4& \ & \textcolor{lightgray}{0}& \ &1\ & \ \\ \ & \ &1& \ &-4& \ &-4& \ &1\ & \ & \ \\ \ & \ & \ &1& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

可以通过把 \(-4\) 拆成两个 \(-2\) 来实现配成两组 \([1,-2,1]\)。留给读者练习。

5.1.1

\(\mathrm{(p(a+b)/8)2-(s(ab)/3)3}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &27& \ &-10& \ &27& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &54& \ &-30& \ &-30& \ &54& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &27& \ &-30& \ &-114& \ &-30& \ &27\ & \ & \ \\ \ & \ & \ &-10& \ &-30& \ &-30& \ &-10\ & \ & \ & \ \\ \ & \ & \ & \ &27& \ &54& \ &27\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

按照正常流程拿掉 \(\mathrm{p(a-b)2}\) 和六芒星剩下

\[\left[\begin{matrix}4& \ &3& \ &3& \ &4\ \\ \ &3& \ &-30& \ &3\ & \ \\ \ & \ &3& \ &3\ & \ & \ \\ \ & \ & \ &4\ & \ & \ & \ \end{matrix}\right] \]

拿掉 Schur 就基本做完了。

5.1.7

\(\mathrm{1/p(a)-s(1/(a3+b3+abc))}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

拿掉 \(\mathrm{s(a6(b+c)(b-c)2)}\) 然后拆成三个 \([1,-1,-1,1]\) 就完事了。

文献 4

3.

\(\mathrm{s((b2/c2+c2/a2)/(a/b))-s(a/b)-3}\)

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-1& \ &-3& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

拿掉 \(\mathrm{s(ab2(ab-c2)2(ab+c2))}\) 之后就剩中间的 \(1\) 围着 \(-3\)。

7.

\(\mathrm{s(a/(b+c)2)-9/4/s(a)}\)

\[\left[\begin{matrix}4& \ &12& \ &3& \ &-10& \ &3& \ &12& \ &4\ \\ \ &12& \ &14& \ &-14& \ &-14& \ &14& \ &12\ & \ \\ \ & \ &3& \ &-14& \ &-30& \ &-14& \ &3\ & \ & \ \\ \ & \ & \ &-10& \ &-14& \ &-14& \ &-10\ & \ & \ & \ \\ \ & \ & \ & \ &3& \ &14& \ &3\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &12& \ &12\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &4\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

常规套路拿掉 \(\mathrm{4s(a4(a-b)(a-c))+16s(ab)s(a2(a-b)(a-c))+19p(a-b)2+28s(ab(a-c)2(b-c)2)}\) 然后降次。

\[\left[\begin{matrix}9& \ &8& \ &8& \ &9\ \\ \ &8& \ &-75& \ &8\ & \ \\ \ & \ &8& \ &8\ & \ & \ \\ \ & \ & \ &9\ & \ & \ & \ \end{matrix}\right] \]

拿掉 Schur 基本就做完了。

28.

\(\mathrm{s((a+b)/(b+c)a/(2a+b+c))-3/4}\)

\[\left[\begin{matrix}4& \ &14& \ &9& \ &-6& \ &1& \ &10& \ &4\ \\ \ &10& \ &18& \ &-14& \ &-22& \ &18& \ &14\ & \ \\ \ & \ &1& \ &-22& \ &-42& \ &-14& \ &9\ & \ & \ \\ \ & \ & \ &-6& \ &-14& \ &-22& \ &-6\ & \ & \ & \ \\ \ & \ & \ & \ &9& \ &18& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &14& \ &10\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &4\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

贪心的扣掉外圈项。\(\mathrm{4s(a4(a-b)(a-c))+14s(ab)s(a2(a-b)(a-c))+15p(a-b)2+4s(ab(a2-c2)2)+24s(ab(a-c)2(b-c)2)+8s(bc2(a+b)(a-b)2)}\)，剩下的除以 \(\mathrm{2p(a)}\).

\[\left[\begin{matrix}15& \ &18& \ &10& \ &15\ \\ \ &10& \ &-129& \ &18\ & \ \\ \ & \ &18& \ &10\ & \ & \ \\ \ & \ & \ &15\ & \ & \ & \ \end{matrix}\right] \]

一眼 Schur 可解，做完了。

29.

\(\mathrm{s(a/b)-s((a+b)/(b+c))}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &1& \ & \textcolor{lightgray}{0}& \ &-3& \ &-1& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &1& \ &-1& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

用 \([1,-2,1]\) 拿掉外圈项之后就是 \(1\) 套 \(-3\) 的情形。

30.

\(\mathrm{s(a3/(b2-bc+c2))-3s(ab)/s(a)}\)

\[\left[\begin{matrix}1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-2& \ &3& \ &-2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1\ \\ \ & \textcolor{lightgray}{0}& \ &-1& \ &1& \ &-1& \ &-1& \ &1& \ &-1& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &1& \ &2& \ &-1& \ &2& \ &1& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &-2& \ &-1& \ &-1& \ &-1& \ &-1& \ &-2\ & \ & \ & \ \\ \ & \ & \ & \ &3& \ &-1& \ &2& \ &-1& \ &3\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &-2& \ &1& \ &1& \ &-2\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &-1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ &1\ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

扣掉 \(\mathrm{s(a4(a2-b2)(a2-c2))}\) 和 \(\mathrm{s(a2)p(a-b)2}\)，剩下比较像六芒星，所以扣线性项 \(\mathrm{1/2s(a4(b-c)4)+1/2s(abc(a+b)(a-b)4)}\)，这样可以降次：

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &4& \ &-6& \ &4& \ &1\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-6& \ &-6& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &4& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

拿掉 \(\mathrm{s(c(a-b)4)}\) 之后只剩中间的 \(\mathrm{s((a-b)2)}\) 了。

35.

\(\mathrm{s(a)/4-s(ab/(a+b+2c))}\)

\[\left[\begin{matrix} 2& \ &1& \ &-6& \ &1& \ &2\ \\ \ &1& \ &2& \ &2& \ &1\ & \ \\ \ & \ &-6& \ &2& \ &-6\ & \ & \ \\ \ & \ & \ &1& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ &2\ & \ & \ & \ & \ \end{matrix}\right] \]

拿掉两个 Schur，边上就剩下 \([3,-6,3]\) 了，结束。

39.

\(\mathrm{s((b+c)/a)-4s(a/(b+c))}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &2& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &-2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-2& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &1& \ & \textcolor{lightgray}{0}& \ &-6& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ \\ \ & \ & \ &2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &-2& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

拿掉 \(\mathrm{p(a-b)2}\) 就可以用 \([1,-1,-1,1]\) 处理掉了。

44.

\(\mathrm{27+p(2+a2/b/c)-6s(a)s(1/a)}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &4& \ &-6& \ &-6& \ &4& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-6& \ &18& \ &-6& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &2& \ &-6& \ &-6& \ &2\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &4& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

六芒星结构，先拿掉。剩下的就是 \(\mathrm{s(a3-abc)}\)。

67.

\(\mathrm{p(a2/b2+2)-9s(a/c)}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ & \textcolor{lightgray}{0}& \ &4& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-9& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &4& \ & \textcolor{lightgray}{0}& \ &9& \ &-9& \ &2\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &-9& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ &2& \ & \textcolor{lightgray}{0}& \ &4\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

先拿掉 \(\mathrm{s(a2(ac-b2)2)}\) 让上方变对称，再按照惯例拿 \(\mathrm{p(a-b)2}\) 和六芒星。剩下的东西没有 \(a^3\) 系数，那么配方就显而易见了。

71.

\(\mathrm{s((a-b)2)/4-s((a3-b3)/(a+b))}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &1& \ &2& \ &-2& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ & \textcolor{lightgray}{0}& \ &-2& \ & \textcolor{lightgray}{0}& \ &1\ & \ \\ \ & \ &-2& \ &-2& \ &-2& \ &2\ & \ & \ \\ \ & \ & \ &2& \ & \textcolor{lightgray}{0}& \ &-2\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

先拿掉 \(\mathrm{s(ab2(a-b)2)}\)，这样外圈负项就没了。剩下的就是三个 \([1,-2,1]\)。

74.

\(\mathrm{s(a2/b2)+5-p(1+a/b)}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-1& \ &-1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &1& \ &-1& \ &3& \ &-1& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &-1& \ &-1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

发现扣什么都会变成负的，所以升次。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-1& \ &-2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &1& \ & \textcolor{lightgray}{0}& \ &1& \ &1& \ &-1& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &1& \ &-2& \ &1& \ &-2& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &-1& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

升完次发现可以减掉 \(\mathrm{1/2s(a3(b2-ac)2)+1/2s(a2c3(a-b)2)}\)，又减不掉了，再升次。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \frac{1}{2}& \ & \frac{1}{2}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-1& \ &-1& \ & \frac{1}{2}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \frac{1}{2}& \ & \frac{1}{2}& \ & \frac{1}{2}& \ & \textcolor{lightgray}{0}& \ & \frac{1}{2}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \frac{1}{2}& \ &-1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &-1& \ & \frac{1}{2}& \ &-1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \frac{1}{2}& \ & \frac{1}{2}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \frac{1}{2}\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

可以看出来就是 \([1,-2,1]\) 多叠了几个。

75.

\(\mathrm{8-s((2a+b+c)2/(2a2+(b+c)2))}\)

\[\left[\begin{matrix}8& \ &4& \ &1& \ &10& \ &1& \ &4& \ &8\ \\ \ &4& \ &10& \ &-26& \ &-26& \ &10& \ &4\ & \ \\ \ & \ &1& \ &-26& \ &42& \ &-26& \ &1\ & \ & \ \\ \ & \ & \ &10& \ &-26& \ &-26& \ &10\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &10& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &4& \ &4\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &8\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

按照惯例拿掉 \(\mathrm{8s(a(a-b)(a-c))2+20s(ab)s(a2(a-b)(a-c))+29p(a-b)+-24s(ab(a-c)2(b-c)2)+12s(a2b2(a-c)(b-c))}\) 就剩下 \(\mathrm{s(a2b+ab2-2abc)}\) 的结构了。

96.

\(\mathrm{s(1/(a2+ab+b2))-9/s(a)2}\)

\[\left[\begin{matrix}1& \ &3& \ &-3& \ &-1& \ &-3& \ &3& \ &1\ \\ \ &3& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &3\ & \ \\ \ & \ &-3& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-3\ & \ & \ \\ \ & \ & \ &-1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-1\ & \ & \ & \ \\ \ & \ & \ & \ &-3& \ & \textcolor{lightgray}{0}& \ &-3\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &3& \ &3\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &1\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

按照常规步骤扣掉 \(\mathrm{s(a2(a2-b2)(a2-c2))+3s(ab)s(a2(a-b)(a-c))+p(a-b)2+s(ab(a-c)2(b-c)2)}\) 剩下可以降次、扣 Schur、结束。

99.

\(\mathrm{s(1/(2+a/b))-s(1/(1+a/b+b/c))}\)

扣掉这些线性项 \(\mathrm{3/2s(ab2(ab-c2)2(ab+c2))+3/2s(a2b5(a-c)2s(ab4(a2-c2)2)}\) 可以降次。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &2& \ &10& \ &10& \ & \textcolor{lightgray}{0}& \ &3& \ & \textcolor{lightgray}{0}\ \\ \ &3& \ &9& \ &-16& \ &-12& \ &9& \ &2\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-12& \ &-18& \ &-16& \ &10\ & \ & \ \\ \ & \ & \ &10& \ &-16& \ &-12& \ &10\ & \ & \ & \ \\ \ & \ & \ & \ &10& \ &9& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &2& \ &3\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

扣 \(\mathrm{2s(ab)s(a2(a-b)(a-c))+s(ab(b2-c2)2)}\)，这样 \(ab^5\) 次数也没了。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &6& \ &5& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &5& \ &-7& \ &-6& \ &5& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &1& \ &-6& \ &-12& \ &-7& \ &6\ & \ & \ \\ \ & \ & \ &5& \ &-7& \ &-6& \ &5\ & \ & \ & \ \\ \ & \ & \ & \ &6& \ &5& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

横向拿掉 \(\mathrm{5s(bc2(a+b)(a-b)2)}\) 之后就是三次了，不多赘述。

102.

\(\mathrm{s((b+c-a)2/((b+c)2+a2))-3/5}\)

\[\left[\begin{matrix}12& \ &4& \ &-4& \ &8& \ &-4& \ &4& \ &12\ \\ \ &4& \ &12& \ &-24& \ &-24& \ &12& \ &4\ & \ \\ \ & \ &-4& \ &-24& \ &48& \ &-24& \ &-4\ & \ & \ \\ \ & \ & \ &8& \ &-24& \ &-24& \ &8\ & \ & \ & \ \\ \ & \ & \ & \ &-4& \ &12& \ &-4\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &4& \ &4\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &12\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

按照常规操作拿掉 \(\mathrm{12s(a4(a-b)(a-c))+16s(ab)s(a2(a-b)(a-c))+12p(a-b)2+32s(ab(a-c)2(b-c)2)}\) 之后就可以降次。

降成三次之后扣掉 Schur 就是 \(\mathrm{s(a2b+ab2-2abc)}\) 了，不多赘述。

108.

证明三元的情况，右边改成 \(\frac34\)。

\(\mathrm{s(1/(1+a/b)2)-3/4}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &2& \ &5& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &2& \ &-2& \ &-2& \ &2& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &5& \ &-2& \ &-18& \ &-2& \ &1\ & \ & \ \\ \ & \ & \ &2& \ &-2& \ &-2& \ &2\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &2& \ &5\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

拿掉 \(\mathrm{p(a-b)2}\) 和六芒星之后剩下这个。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ &2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &1& \ &2& \ &-15& \ &2& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &2& \ &2& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

再拿掉横向的 \(\mathrm{s(ac2(a+b)(a-b)2)}\) 就可以降次，不再赘述。

109.

\(\mathrm{s(a2/(b2+c2))-s(a/(b+c))}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-1& \ &-1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &2& \ &1& \ &-1& \ & \textcolor{lightgray}{0}& \ &-1& \ &1& \ &2& \ &1\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &1& \ &-2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-2& \ &1& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &-1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ &-1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &-1& \ &-1& \ &-2& \ &-1& \ &-1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &1& \ &1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &2& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

拿掉外圈的 \(\mathrm{s(ab(a3-b3)(a4-b4))}\) 就可以降次。

\[\left[\begin{matrix}2& \ &1& \ &-1& \ & \textcolor{lightgray}{0}& \ &-1& \ &1& \ &2\ \\ \ &1& \ &-2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-2& \ &1\ & \ \\ \ & \ &-1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &-1\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ &-1& \ &-2& \ &-1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &2\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

拿掉 \(\mathrm{s((a2+b2)(a2-b2)2)}\) 就是三个 \([1,-2,1]\) 了。

一些杂项

1

\(\mathrm{s(a6)-3p(a)2-1/2p(a-b)2}\)

\[\left[\begin{matrix}2& \ & \textcolor{lightgray}{0}& \ &-1& \ &2& \ &-1& \ & \textcolor{lightgray}{0}& \ &2\ \\ \ & \textcolor{lightgray}{0}& \ &2& \ &-2& \ &-2& \ &2& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &-1& \ &-2& \ & \textcolor{lightgray}{0}& \ &-2& \ &-1\ & \ & \ \\ \ & \ & \ &2& \ &-2& \ &-2& \ &2\ & \ & \ & \ \\ \ & \ & \ & \ &-1& \ &2& \ &-1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &2\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

用 \(\mathrm{s((a2+b2)(a2-b2)2)}\) 拿掉最外层的负项。剩下的可以用两个 \([1,-1,-1,1]\) 扣完。

2

\(\mathrm{(s(a2)/3)3-p(a)2-1/16p(a-b)2}\)

\[\left[\begin{matrix}16& \ & \textcolor{lightgray}{0}& \ &21& \ &54& \ &21& \ & \textcolor{lightgray}{0}& \ &16\ \\ \ & \textcolor{lightgray}{0}& \ &54& \ &-54& \ &-54& \ &54& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &21& \ &-54& \ &-174& \ &-54& \ &21\ & \ & \ \\ \ & \ & \ &54& \ &-54& \ &-54& \ &54\ & \ & \ & \ \\ \ & \ & \ & \ &21& \ &54& \ &21\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &16\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

外层用 \(\mathrm{16s(a2(a2-b2)(a2-c2))}\) 先拿掉角，再用 \(\mathrm{p(a-b)2}\)，就和上一题一样了。

3

\(\mathrm{s(a2)3-27p(a)2-2p(a-b)2}\)

\[\left[\begin{matrix}1& \ & \textcolor{lightgray}{0}& \ &1& \ &4& \ &1& \ & \textcolor{lightgray}{0}& \ &1\ \\ \ & \textcolor{lightgray}{0}& \ &4& \ &-4& \ &-4& \ &4& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &1& \ &-4& \ &-9& \ &-4& \ &1\ & \ & \ \\ \ & \ & \ &4& \ &-4& \ &-4& \ &4\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &4& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &1\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

先拿这个 \(\mathrm{s(a2(a2-b2)(a2-c2))}\)，让它变成可以减 \(\mathrm{p(a-b)2}\) 的，然后减掉，就和上一题一样了。

4

\(\mathrm{8s(a2(b2-c2)2)-3p(a-b)2}\)

拿掉 \(\mathrm{p(a-b)2}\) 之后用前面的做法就好了。

5

\(\mathrm{p(a2+ab+b2)-3s(ab)s(a2b2)}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &-2& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &1& \ &-1& \ &-1& \ &1& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &1& \ &-1& \ &3& \ &-1& \ &1\ & \ & \ \\ \ & \ & \ &-2& \ &-1& \ &-1& \ &-2\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &1& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

拿掉 \(\mathrm{p(a-b)2}\) 之后就是 Schur 了。

6

\(\mathrm{(s(a3)+3abc)/s(ab(a+b))+5/4-s(ab)s(1/(a+b)2)}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ &-1& \ &-1& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &7& \ &-2& \ &-10& \ &-2& \ &7& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &7& \ &6& \ &-5& \ &-5& \ &6& \ &7& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &1& \ &-2& \ &-5& \ &12& \ &-5& \ &-2& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ &-1& \ &-10& \ &-5& \ &-5& \ &-10& \ &-1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &-1& \ &-2& \ &6& \ &-2& \ &-1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &1& \ &7& \ &7& \ &1\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

显然先拿掉 \(\mathrm{s(ab)s(a)p(a-b)2}\)，这样可以降次。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &8& \ &-3& \ &-10& \ &-3& \ &8& \ & \textcolor{lightgray}{0}\ \\ \ &8& \ &6& \ &-6& \ &-6& \ &6& \ &8\ & \ \\ \ & \ &-3& \ &-6& \ &18& \ &-6& \ &-3\ & \ & \ \\ \ & \ & \ &-10& \ &-6& \ &-6& \ &-10\ & \ & \ & \ \\ \ & \ & \ & \ &-3& \ &6& \ &-3\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &8& \ &8\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

再拿掉 \(\mathrm{8s(ab)s(a2(a-b)(a-c))+5p(a-b)2}\) 变成三次，Schur 扣掉之后就是 \(\mathrm{s(a(b-c)2)}\)。

7

\(\mathrm{s(a3/b3)+6-3s(b/a)}\)

结构太稀疏，这里就不放了。

由于这个不等式很松，可以直接拿掉 90 度躺着的 Schur 降次。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &1& \ &-3& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &3& \ &-3& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &1& \ &-3& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

拿掉六芒星之后就是三次了。

8

\(\mathrm{s(1/(1+a/b)3)+5/p(1+a/b)-1}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ &1& \ &-1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ &-3& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ &1& \ &-1& \ &-3& \ &-3& \ & \textcolor{lightgray}{0}& \ &2& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ &-3& \ &1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &2& \ &1& \ &-1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &1\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

用 \(\mathrm{s(a3b3(b+c)(b-c)2)}\) 拿掉外圈项，然后降次。

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ &1& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &1& \ & \textcolor{lightgray}{0}& \ &-2& \ &1& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ & \textcolor{lightgray}{0}& \ &-2& \ &-6& \ & \textcolor{lightgray}{0}& \ &2\ & \ & \ \\ \ & \ & \ &1& \ & \textcolor{lightgray}{0}& \ &-2& \ &1\ & \ & \ & \ \\ \ & \ & \ & \ &2& \ &1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

把 \(2,-2\) 单独放一组，\(1,-6\) 单独放一组，都是比较容易理解的结构。

9

\(\mathrm{21s(a4)+s(a)4-3s((a+b)4)}\)

拿掉 Schur 之后用 \([2,-4,2]\) 减一下就好了。

10

\(\mathrm{(s(1/(a+b))+3/s(a))2-16/s(ab)}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ &1& \ &-2& \ &-1& \ &4& \ &-1& \ &-2& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &-3& \ &13& \ &69& \ &69& \ &13& \ &-3& \ &1\ & \ \\ \ & \ &-2& \ &13& \ &132& \ &235& \ &132& \ &13& \ &-2\ & \ & \ \\ \ & \ & \ &-1& \ &69& \ &235& \ &235& \ &69& \ &-1\ & \ & \ & \ \\ \ & \ & \ & \ &4& \ &69& \ &132& \ &69& \ &4\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &-1& \ &13& \ &13& \ &-1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &-2& \ &-3& \ &-2\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

外圈刚好是 \([1,-2,-1,4,-1,-2,1]\)，刚好可以直接用 \(\mathrm{s(ab)s(a(a-b)(a-c))2}\) 拿完。拿掉之后剩下的全部正，结束。

11

\(\mathrm{s(1/(a3+b3+abc))+1/p(a)-54/s(a)3}\)

乱拿 \(\mathrm{s(a3b3)s(a4(a-b)(a-c))+4s(a4b4)s(a2(a-b)(a-c))+7s(a7(b+c)(b-c)4)+2s(ab(ab-c2)4(ab+c2))}\) 之后成功降次。

剩下的先放着。

12

\(\mathrm{s(a/b)6-27p(a3/b3+2)}\)

转置一下可以变成一个六次不等式。

\[\left[\begin{matrix}1& \ &6& \ &15& \ &-34& \ &15& \ &6& \ &1\ \\ \ &6& \ &-78& \ &60& \ &60& \ &-78& \ &6\ & \ \\ \ & \ &15& \ &60& \ &-153& \ &60& \ &15\ & \ & \ \\ \ & \ & \ &-34& \ &60& \ &60& \ &-34\ & \ & \ & \ \\ \ & \ & \ & \ &15& \ &-78& \ &15\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &6& \ &6\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &1\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

照例 \(\mathrm{s(a(a-b)(a-c))2+2s(ab)s((a-b)2)2}\) 拿掉 Schur。再拿掉对应的 \(\mathrm{p(a-b)2}\) 发现只剩六芒星了。

13

\(\mathrm{s((s(ab)2+a2b2)/(a+b)2)-5/2s(ab)}\)

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ &-1& \ &-2& \ &-1& \ &2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &4& \ &1& \ &-5& \ &-5& \ &1& \ &4& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &2& \ &1& \ &-2& \ &6& \ &-2& \ &1& \ &2\ & \ & \ \\ \ & \ & \ &-1& \ &-5& \ &6& \ &6& \ &-5& \ &-1\ & \ & \ & \ \\ \ & \ & \ & \ &-2& \ &-5& \ &-2& \ &-5& \ &-2\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &-1& \ &1& \ &1& \ &-1\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ &2& \ &4& \ &2\ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

按照常规方法，扣掉 \(\mathrm{2s(a2b2)s(a2(a-b)(a-c))+s(ab)p(a-b)2}\)，可以降次。

\[\left[\begin{matrix}2& \ &2& \ &-4& \ &-4& \ &2& \ &2\ \\ \ &2& \ &-3& \ &5& \ &-3& \ &2\ & \ \\ \ & \ &-4& \ &5& \ &5& \ &-4\ & \ & \ \\ \ & \ & \ &-4& \ &-3& \ &-4\ & \ & \ & \ \\ \ & \ & \ & \ &2& \ &2\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ &2\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

按照常规方法拿掉 \(\mathrm{2s(a(a-b)2(a-c)2)}\)，五次下还可以考虑 \(\mathrm{p(a)s((a-b)(a-c))}\)。除以 \(6\) 就是这个：

\[\left[\begin{matrix} \textcolor{lightgray}{0}& \ &1& \ &-1& \ &-1& \ &1& \ & \textcolor{lightgray}{0}\ \\ \ &1& \ &-2& \ &2& \ &-2& \ &1\ & \ \\ \ & \ &-1& \ &2& \ &2& \ &-1\ & \ & \ \\ \ & \ & \ &-1& \ &-2& \ &-1\ & \ & \ & \ \\ \ & \ & \ & \ &1& \ &1\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \end{matrix}\right] \]

这个结构仍然需要背下来：\(\mathrm{s(a(b-c)2(b+c-a)2)}\)。

14

\(\mathrm{s(1/(1+a/b))-s(3/(4+a/b+b/a))}\)

先用 \(\mathrm{s(a3b3)s(a(a-b)(a-c))+3s(a2b(ab+c2)(ab-c2)2)+6s(ab2c2(a2-c2)2)+s(b4(a+c)(a-c)4)+s(a3b(b+c)(b-c)4)+6s(b4c(a2-c2)2)}\) 降次。

继续按照常规套路拿 \(\mathrm{6p(a-b)2+21s(ab(a-c)2(b-c)2)+16s(ac2(a+b)(a-b)2)+3s(a2b2(a-c)(b-c))}\) 就降到三次了。

15

\(\sum_{\mathrm{cyc}}\frac{a^2b^2}{c^2}\ge\sum_{\mathrm{cyc}}\frac{b^2c}a\)

\(\mathrm{s(b4c4-ab4c3)}\)

系数阵过于显然，不给了。显然可以通过 \(\frac1a,\frac1b,\frac1c\) 把它变成七次，且变出来的七次可以降次。这里结果是 \(\mathrm{s(a4-ab3)}\)，用三元四次定理即可。当然，拿掉 Schur 之后剩下的是 \([1,-2,1]\)，也可以做。

16

\(\mathrm{s(a/b)-s(a/b/(1+a/c))-3/2}\)

\[\left[\begin{matrix}\textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}& \ &2& \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ \\ \ & \textcolor{lightgray}{0}& \ &2& \ &-1& \ &-1& \ &2& \ & \textcolor{lightgray}{0}\ & \ \\ \ & \ &2& \ &-1& \ &-6& \ &-1& \ & \textcolor{lightgray}{0}\ & \ & \ \\ \ & \ & \ & \textcolor{lightgray}{0}& \ &-1& \ &-1& \ & \textcolor{lightgray}{0}\ & \ & \ & \ \\ \ & \ & \ & \ & \textcolor{lightgray}{0}& \ &2& \ &2\ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \textcolor{lightgray}{0}& \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ \\ \ & \ & \ & \ & \ & \ & \textcolor{lightgray}{0}\ & \ & \ & \ & \ & \ & \ \end{matrix}\right] \]

考虑用一种叫 \(\mathrm{s(ab(b-c))2}\) 的结构降次。这个结构最外圈只有 \(a^4b^2\) 是 \(1\)，其它都是 \(0\)。降次之后就没东西了。