$a,b,c$ are real positive numbers ,what is the proof that :
$\frac{2a}{3a^2+b^2+2ac} +\frac{2b}{3b^2+c^2+2ab}+\frac{2c}{3c^2+a^2+2bc}\le\frac{3}{a+b+c}$
$a,b,c$ are real positive numbers ,what is the proof that :
$\frac{2a}{3a^2+b^2+2ac} +\frac{2b}{3b^2+c^2+2ab}+\frac{2c}{3c^2+a^2+2bc}\le\frac{3}{a+b+c}$
Hint: Prove that $ \frac{2a}{3a^2+b^2+2ac}\leq\frac{1}{a+b+c} $
I just follow the very good hint from Norbert's answer
By AM-GM $\sum_{cyc}\frac{2a}{3a^2+b^2+2ac}=\sum_{cyc}\frac{2a}{2a^2+a^2+b^2+2ac}\leq\sum_{cyc}\frac{2a}{2a^2+2ab+2ac}=\frac{3}{a+b+c}$