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[不等式] 证$\frac{z^2-x^2}{x+y}+\frac{x^2-y^2}{y+z}+\frac{y^2-z^2}{x+z}\geqslant 0$

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isee Posted at 2025-3-2 17:44:02 |Read mode
源自知乎提问





展开,高联一试的基操吧




分析法,对于正实数 x,y,z 有 \begin{gather*}
\frac{z^2-x^2}{x+y}+\frac{x^2-y^2}{y+z}+\frac{y^2-z^2}{x+z}\geqslant 0\\[1ex]
\iff \frac{x^4+y^4+z^4-x^2y^2-y^2z^2-z^2x^2}{(x+y) (x+z) (y+z)}\geqslant 0\\[1ex]
\iff x^4+y^4+z^4\geqslant x^2y^2+y^2z^2+z^2x^2,\\

\end{gather*} 这是成立的,因为 \[\left\{ \begin{aligned} x^4+y^4&\geqslant 2x^2y^2,\\[1ex]
y^4+z^4&\geqslant 2y^2z^2,\\[1ex]
z^4+x^4&\geqslant 2z^2x^2, \end{aligned} \right.\] 三式相加整理即证.
isee=freeMaths@知乎

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2025-4-20 22:18 GMT+8

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