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[不等式] 三元不等式的证明有一步不知是怎么得来的

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hbghlyj Posted 2021-4-4 12:55 |Read mode

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a,b,c>0,则${\sqrt[3]{abc}(2a+2b+2c+ab+bc+ca)}\leq {3abc+2\sum a^{2}}$

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kuing Posted 2021-4-4 13:36

此不等式不成立，反例：a=b=8, c=1

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Author| hbghlyj Posted 2021-4-4 22:02

突然发现,他第一步就放缩过度了

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