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[不等式] 三元不等式

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hbghlyj posted 2021-5-14 16:39 |Read mode
a,b,c>0,$k\ge\frac1{\sqrt2}$,证明$k(a^2 + b^2 + c^2) + abc + 3k + 2 \geq (2k + 1)(a + b + c)$.
当a=b=c时可以取等.
当$k=\frac1{\sqrt2}$时$a = b = 1 + \frac{1}{\sqrt{2}}, c = 0$可以取等.

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