据说爆难

天音

$x, y, z$ 为正数，求证：
\[
(x^2+y^2+z^2)\left(\frac{x^2}{\left(x^2+y z\right)^2}+\frac{y^2}{\left(y^2+z x\right)^2}+\frac{z^2}{\left(z^2+x y\right)^2}\right) \geq \frac{9}{4}
\]

天音

顶下，有人吗

Infinity

这道题至今没能解决吗？

hbghlyj

连续函数$f(x,y,z)=\frac{x^2}{\left(x^2+y z\right)^2}+\frac{y^2}{\left(y^2+z x\right)^2}+\frac{z^2}{\left(z^2+x y\right)^2}$在紧集$x^2+y^2+z^2=1$上必有最大值和最小值。
计算Lagrange multiplier $f+\lambda(x^2+y^2+z^2-1)$的驻点\[
\begin{cases}
\frac{\partial f}{\partial x}=2 \lambda x,\\[1mm]
\frac{\partial f}{\partial y}=2 \lambda y,\\[1mm]
\frac{\partial f}{\partial z}=2 \lambda z,\\[1mm]
x^2+y^2+z^2=1
\end{cases}
\]解得
$\{\lambda = -4,\quad |x| = \sqrt{2}/2,\quad |y| = \sqrt{2}/2,\quad |z| = 0\},$
$\{\lambda = -4,\quad |x| = \sqrt{2}/2,\quad |y| = 0,\quad |z| = \sqrt{2}/2\},$
$\{\lambda = -4,\quad |x| = 0,\quad |y| = \sqrt{2}/2,\quad |z| = \sqrt{2}/2\},$
$\{\lambda = -\frac94,\quad |x| = \sqrt{1/3},\quad |y| = \sqrt{1/3},\quad |z| = \sqrt{1/3}\}$
因此函数$f$在$x^2+y^2+z^2=1$上的最大值为$f(\frac1{\sqrt2},\frac1{\sqrt2},0)=4$，最小值为$f(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3})=\frac94$.