[不等式] 谁能解释下Can是怎么配的？

前天在Aops 看到一个题，如下：
For $a,b,c \in \mathbb{R}$ for which $(a-b)(a-c)(b-c) \neq 0$, prove:

$\sum \frac{a^4}{(a-b)^2(a-c)^2} \ge \frac{1}{2}$

Equality holds when $\{a,b,c\}=\{t,-t,0\}$ for $ t\in \mathbb{R} \backslash \{0\}$

Can给出了他的解答：
Proof.
Using the Cauchy-Schwarz inequality, we have
\[\left[ \sum (2a-b-c)^4\right] \left[ \sum \frac{a^4}{(a-b)^2(a-c)^2}\right] \ge \left[ \sum \frac{a^2(2a-b-c)^2}{(a-b)(a-c)}\right]^2.\]
But, it is easy to check that
\[\sum (2a-b-c)^4 =18\left( \sum a^2-\sum ab\right)^2\]
and
\[\begin{aligned}\sum \frac{a^2(2a-b-c)^2}{(a-b)(a-c)} &=4\sum a^2-\sum ab =\left( \sum a\right)^2+3\left( \sum a^2-\sum ab\right) \\ &\ge 3\left( \sum a^2-\sum ab\right) > 0.\end{aligned} \] So we have \[18\left( \sum a^2-\sum ab\right)^2\left[ \sum \frac{a^4}{(a-b)^2(a-c)^2}\right]  \ge \left[ 3\left( \sum a^2-\sum ab\right)\right]^2,\] from which it follows that \[\sum \frac{a^4}{(a-b)^2(a-c)^2} \ge \frac{1}{2},\] as desired. $\blacksquare$


我想问那个$ \sum (2a-b-c)^4 =18\left( \sum a^2-\sum ab\right)^2$ 是怎么想到的？
问Can, 他说:it's my secret,so I can't tell you now....

Any idea?

我想可能是他早就对四次的恒等式研究得很多了，比如可能推过这个
\begin{align*}
\sum{(x-y)^{4}}&=\sum{\left( (x-y)^{2} \right)^{2}} \\
& =\left( \sum{(x-y)^{2}} \right)^{2}-2\sum{(x-y)^{2}(y-z)^{2}} \\
& =\left( 2\sum{(x-y)(y-z)} \right)^{2}-2\sum{(x-y)^{2}(y-z)^{2}} \\
& =4\left( \sum{(x-y)(y-z)} \right)^{2}-2\left( \sum{(x-y)(y-z)} \right)^{2}+4(x-y)(y-z)(z-x)\sum{(x-y)} \\
& =2\left( \sum{(x-y)(y-z)} \right)^{2} \\
& =\frac{1}{2}\left( \sum{(x-y)^{2}} \right)^{2},
\end{align*}
这样若令 $x=a-b,y=b-c,z=c-a$，则 $x+y+z=0$ 且 $\sum{(x-y)^{4}}=\sum{(2a-b-c)^{4}}$，以及
\begin{align*}
\frac{1}{2}\left( \sum{(x-y)^{2}} \right)^{2}&=2\left( \left( \sum{x} \right)^{2}-3\sum{xy} \right)^{2} \\
& =18\left( \sum{xy} \right)^{2} \\
& =18\left( \sum{(a-b)(b-c)} \right)^{2} \\
& =18\left( \sum{a^{2}}-\sum{ab} \right)^{2}.
\end{align*}