proving :$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$. Let $a,b,c>0$ how to prove that :
$$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$$
I find that
$$\ \frac{ab}{a^{2}+3b^{2}}=\frac{1}{\frac{a^{2}+3b^{2}}{ab}}=\frac{1}{\frac{a}{b}+\frac{3b}{a}} $$
By AM-GM
$$\ \frac{ab}{a^{2}+3b^{2}} \leq \frac{1}{2 \sqrt{3}}=\frac{\sqrt{3}}{6} $$
$$\ \sum_{cyc} \frac{ab}{a^{2}+3b^{2}} \leq \frac{\sqrt{3}}{2} $$
But this is obviously is not working .

From the question it is obvious that equality is achieved when $a=b=c$, this hints at the use of AM-GM. Firstly, use the fact that $ab \le \frac{1}{2} (a^2+b^2)$, we just need to show 
$$ \frac{a^2+b^2}{a^2+3b^2} + \frac{b^2+c^2}{b^2+3c^2} + \frac{c^2+a^2}{c^2+3a^2} \le \frac{3}{2},$$ which is equivalent to (after removing a constant 1 from each term in LHS)
$$ \frac{b^2}{a^2+3b^2} + \frac{c^2}{b^2+3c^2} + \frac{a^2}{c^2+3a^2} \ge \frac{3}{4},$$
by AM-GM, $a^2 + 3b^2 \ge 4 \sqrt[4]{a^2b^6} = 4\sqrt{ab^3}$, hence we just need
$$ \frac{\sqrt{b}}{\sqrt{a}} + \frac{\sqrt{c}}{\sqrt{b}} + \frac{\sqrt{a}}{\sqrt{c}} \ge 3, $$ but this is obviously true by one more application of AM-GM.
