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 .

I have a Cauchy-Schwarz proof of it,hope you enjoy.:D
first,mutiply $2$ to each side,your inequality can be rewrite into
$$ \sum_{cyc}{\frac{2ab}{a^2+3b^2}}\leq \frac{3}{2}$$
Or
$$ \sum_{cyc}{\frac{(a-b)^2+2b^2}{a^2+3b^2}}\geq \frac{3}{2}$$
Now,Using Cauchy-Schwarz inequality,we have
$$ \sum_{cyc}{\frac{(a-b)^2+2b^2}{a^2+3b^2}}\geq \frac{\left(\sum_{cyc}{\sqrt{(a-b)^2+2b^2}}\right)^2}{4(a^2+b^2+c^2)}$$
Therefore,it's suffice to prove
$$\left(\sum_{cyc}{\sqrt{(a-b)^2+2b^2}}\right)^2\geq 6(a^2+b^2+c^2) $$
after simply expand,it's equal to
$$ \sum_{cyc}{\sqrt{[(a-b)^2+2b^2][(b-c)^2+2c^2]}}\geq a^2+b^2+c^2+ab+bc+ca $$
Now,Using Cauchy-Schwarz again,we notice that
$$ \sqrt{[(a-b)^2+2b^2][(b-c)^2+2c^2]}\geq (b-a)(b-c)+2bc=b^2+ac+bc-ab$$
sum them up,the result follows.
Hence we are done!
Equality occurs when $a=b=c$
