# 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 $$\frac{2a}{3a^2+b^2+2ac}\leq\frac{1}{a+b+c}$$
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