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[不等式] $abc=1$

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v6mm131 Posted at 2017-8-28 17:37:01 |Read mode
Last edited by hbghlyj at 2025-4-10 00:58:19题1:已知$a,b,c>0,abc=1$证明:\[a^{3}+b^{3}+c^{3}+\frac{ab}{a^{2}+b^{2}}+\frac{bc}{b^{2}+c^{2}}+\frac{ac}{c^{2}+a^{2}}\geq \frac{9}{5}(a^2+b^2+c^2-\frac{1}{2})\]

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 Author| v6mm131 Posted at 2017-8-29 07:20:33
题2:已知$a,b,c>0,abc=1$证明:\[\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} +3 \geq 2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\]

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 Author| v6mm131 Posted at 2017-8-29 07:23:02
题3:已知$a,b,c>0,abc=1$证明:\[a^{3}b+b^{3}c+c^{3}a +15 \geq 6(a+b+c)\]

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2025-4-20 22:22 GMT+8

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