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$a,b,c$ are real positive numbers ,what is the proof that :

$\frac{2a}{3a^2+b^2+2ac} +\frac{2b}{3b^2+c^2+2ab}+\frac{2c}{3c^2+a^2+2bc}\le\frac{3}{a+b+c}$

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Hint: Prove that $ \frac{2a}{3a^2+b^2+2ac}\leq\frac{1}{a+b+c} $

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    Additional hint if OP is unsatisfied: Note that $2ab\leq a^2+b^2$ since squares are always positive, that is $0\leq (a-b)^2$.2012-08-30
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I just follow the very good hint from Norbert's answer answer

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    Here you can find [complete faq](http://meta.math.stackexchange.com/questions/107/faq-for-math-stackexchange/117#117) on writing formulas. And please, don't post complete solutions.2012-08-30
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By AM-GM $\sum_{cyc}\frac{2a}{3a^2+b^2+2ac}=\sum_{cyc}\frac{2a}{2a^2+a^2+b^2+2ac}\leq\sum_{cyc}\frac{2a}{2a^2+2ab+2ac}=\frac{3}{a+b+c}$