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[不等式] 关于三元均值不等式

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hbghlyj Posted 2020-4-17 15:21 |Read mode
Last edited by hbghlyj 2020-4-19 09:36设a,b,c为任意正数,A为其算术平均数,G为几何平均数,m为实数,则
(1)$(a^3+b^3+c^3)G≥3Aabc$
(2)(舒尔不等式)$a^3+b^3+c^3-3abc≥a(b-c)^2+b(c-a)^2+c(a-b)^2$
(3)$a^3+b^3+c^3\ge\frac{(a+b+c)^3}9\ge\frac38(a+b)(b+c)(c+a)\ge3abc$
(4)$a^3+b^3+c^3\ge 4a^mb^mc^m-a^{4m-3}b^{4m-3}c^{4m-3}$
(5)$a^3+b^3+c^3-3 a b c\geq \sqrt{3 \left(2 \sqrt{3}+3\right)} |(a-b) (a-c) (b-c)|$当$a=b,c=0$时取等

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kuing Posted 2020-4-18 15:23
(5)《撸题集》P.467 题目 4.6.46.

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