四元四次对称实数不等式

yao4015

$x,y,z,w$ 是实数. 证明
\[
x(x-y)(x-z)(x-w)+y(y-x)(y-z)(y-w)+z(z-x)(z-y)(z-w)+w(w-x)(w-y)(w-z)+xyzw\geq 0.
\]

kuing

作置换 `(x,y,z,w)\to (a_1-a_5,a_2-a_5,a_3-a_5,a_4-a_5)`，不等式变成
\begin{align*}
&(a_1-a_2)(a_1-a_3)(a_1-a_4)(a_1-a_5)\\
&+(a_2-a_1)(a_2-a_3)(a_2-a_4)(a_2-a_5)\\
&+(a_3-a_1)(a_3-a_2)(a_3-a_4)(a_3-a_5)\\
&+(a_4-a_1)(a_4-a_2)(a_4-a_3)(a_4-a_5)\\
&+(a_5-a_1)(a_5-a_2)(a_5-a_3)(a_5-a_4)\geqslant0,
\end{align*}
在数轴上以 `A_i` 表示数值 `a_i` 对应的点，由对称性不妨设 `A_1` 至 `A_5` 在数轴上顺次排列，则不等式就是
\begin{align*}
&|A_1A_2|\cdot|A_1A_3|\cdot|A_1A_4|\cdot|A_1A_5|\\
&-|A_2A_1|\cdot|A_2A_3|\cdot|A_2A_4|\cdot|A_2A_5|\\
&+|A_3A_1|\cdot|A_3A_2|\cdot|A_3A_4|\cdot|A_3A_5|\\
&-|A_4A_1|\cdot|A_4A_2|\cdot|A_4A_3|\cdot|A_4A_5|\\
&+|A_5A_1|\cdot|A_5A_2|\cdot|A_5A_3|\cdot|A_5A_4|\geqslant0,
\end{align*}
显然上式第一项 `\geqslant` 第二项，第五项 `\geqslant` 第四项，所以不等式成立。

isee

好奇怪，我联想到复数了