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

$a,b,c$ 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 $0\leq (a-b)^2$. –  Eric Naslund Aug 30 '12 at 2:28

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Apart from the fact that you wrote the proof backwards without saying so (instead of the ususal "we know X therefore Y and we can conclude Z" your proof should read "we want to prove Z which will follow if Y for which it will suffice that X which is true), it is also unusual to post jour answer as jpeg file. –  Marc van Leeuwen Aug 30 '12 at 6:34
Sorry about posting my answer as jpg file. I am a new user of this website. I had been trying to write equations in my post, but I didn't know how to, hence I had to write in MS Word then captured it. –  Duc Trung Aug 30 '12 at 6:49
Here you can find complete faq on writing formulas. And please, don't post complete solutions. –  Norbert Aug 30 '12 at 6:58