Last edited by hbghlyj at 2025-4-9 07:42:59设 $x, y, z, a, b, c$ 为非负数 $x \geq a, z \leq c$．求证
\[
\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}+\frac{y^2}{x}+\frac{z^2}{y}+\frac{x^2}{z} \geq \sqrt{6\left(a^2+b^2+c^2+x^2+y^2+z^2\right)}
\]
Wanhuihua 20170306
