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Last edited by hbghlyj 2025-3-21 19:06设 $x, y, z \inR^*$，求证：$$\frac{(x+1)(y+1)^2}{3 \sqrt[3]{z^2 x^2}+1}+\frac{(y+1)(z+1)^2}{3 \sqrt[3]{x^2 y^2}+1}+\frac{(z+1)(x+1)^2}{3 \sqrt[3]{y^2 z^2}+1} \geqslant x+y+z+3$$
分母三次根号神马的明显只是唬人的……
\begin{align*}
\sum\frac{(x+1)(y+1)^2}{3\sqrt[3]{z^2x^2}+1}&\geqslant\sum\frac{(x+1)(y+1)^2}{zx+z+x+1} \\
& =\sum\frac{(y+1)^2}{z+1} \\
& \geqslant \frac{\left( \sum(y+1) \right)^2}{\sum(z+1)} \\
& =x+y+z+3.
\end{align*} |
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