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 answered this question in a slightly different way. 
Let us assume the following : $ \frac{a}{b}=x$ and $ \frac{b}{c}=y.$ This converts the above equation to a equation 
with two variables. $$ f(x,y)=\frac{x}{3+x^2} + \frac{y}{3+y^2}+\frac{xy}{1+3(xy)^2}$$ Now to get a maxima or minima point of $f(x,y)$ we partially differentiate it with $x$ and $y$ and equate them to $0$. Hence we have $$ \frac{\partial f(x,y)}{\partial x}= \frac{3-x^2}{(x^2+3)^2}+\frac{y(1-3x^2y^2)}{(3x^2y^2+1)^2}=0 $$ $$\Rightarrow \frac{3-x^2}{y(x^2+3)^2}=\frac{(3x^2y^2-1)}{(3x^2y^2+1)^2}...........Eqn(1)$$ Since $f(x,y)$ is symmetric with $x$ and $y$ ,we also have for $\frac{\partial f(x,y)}{\partial y}=0$ $$\frac{3-y^2}{x(y^2+3)^2}=\frac{(3x^2y^2-1)}{(3x^2y^2+1)^2}...........Eqn(2)$$ Combining equation 1 and 2 we get $$x \cdot\frac{3-x^2}{(x^2+3)^2}=y \cdot\frac {3-y^2}{(y^2+3)^2}$$ This immediately shows that $x=y $ is a critical point (maxima or minima). Now this clearly shows that at the critical point $$x=y$$ $$\Rightarrow ac=b^2.........Eqn (3)$$ In a similar fashion assuming $\frac{b}{c}=x$ and $\frac{c}{a}=y$ we again the same set of equation $$ f(x,y)=\frac{x}{3+x^2} + \frac{y}{3+y^2}+ \frac{xy}{1+3(xy)^2}$$ Following the same steps we get $$x=y$$ $$\Rightarrow ab=c^2...........Eqn(4)$$ Equation 3 and 4 show that $$a=b=c$$ at the critical point. Hence $a=b=c$ gives us that $$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}=\frac{3}{4}$$ is a maxima or a minima. To check maxima we double differentiate and check $\frac{\partial^2f}{\partial x^2}$ and $\frac{\partial^2f}{\partial y^2}$ We get the following : $$\frac{\partial^2f}{\partial x^2}=\frac{\partial^2f}{\partial y^2}=\frac{2x^3-18x}{(3+x^2)2}+\frac{18x^4-18x^8}{(1+3x^4)^2}=-\frac{16}{64} $$ at $ x=y=1$. Both being negative we see that the function $f$ has a maxima at $a=b=c$ which is $\frac{3}{4}$ Hence $$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2} \le \frac{3}{4}$$
