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[不等式] 再一道不等式

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longzaifei Posted at 2014-1-12 14:21:24 |Read mode
Last edited by longzaifei at 2014-1-12 14:47:00设$a,b,c>0$,\[ \frac{a^2}{b(a^2-ab+b^2)}+\frac{b^2}{c(b^2-bc+c^2)}+\frac{c^2}{a(c^2-ca+a^2)}\ge\frac{9}{a+b+c} \]

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kuing Posted at 2014-5-4 17:37:01
难  顶一下

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realnumber Posted at 2014-5-4 20:49:00
$type 一个不等式9÷(a b c).gsp (3.1 KB, Downloads: 2593) 回复 2# kuing
几何画板实验了下,发现左边大于等于$\frac{2}{a+b}+\frac{2}{a+c}+\frac{2}{b+c}$.
证明不会.

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其妙 Posted at 2014-5-4 23:39:55
回复 3# realnumber
涨姿势了,real太厉害了!

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