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 \(p(a)\) 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。
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