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[不等式] 关于$\frac1{a+3}$的四元不等式

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hbghlyj Posted 2021-4-4 21:56 |Read mode
a,b,c,d>0,$a^2+b^2+c^2+d^2=4$,求证$1\le\sum\frac{1}{a+3}\le\frac65$.

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kuing Posted 2021-4-4 22:04
这个没难度,皆因 `\frac1{\sqrt x+3}` 是下凸的。

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