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[不等式] 软件验证下是否成立就可以,不需要证明

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realnumber Posted 2014-4-1 19:57 |Read mode
没装不等式软件,网友要求验证下
$a,b,c\in R+,a+b+c=1$,以下不等式是否成立$4(a^2b^2+c^2b^2+a^2c^2)+6abc(a^2+b^2+c^2)\le a^3b+ab^3+a^3c+c^3a+b^3c+c^3b+4abc$.

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kuing Posted 2014-4-1 21:25
当 b=a, c=1-2a 时 右减左 = 2a(4a-1)(a-1)(3a-1)^2,显然不成立。

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