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[不等式] 好久没发帖了，来个三元不等式

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Shiki posted 2019-11-25 22:24 |Read mode

似乎不太难...

$$\sum \sqrt{a^2-ab+b^2} +9\sqrt[3]{abc} \leqslant 4(a+b+c) $$

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original poster Shiki posted 2019-12-1 13:22

不能说不太难，是相当简单

$$LHS \leqslant \sum \sqrt{a^2-ab+b^2}+3\sum \sqrt{ab}$$

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2025-7-15 15:27 GMT+8

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