proving :$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$.

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 $a=b=c$

(2018-01-01) Ahhh, so sweet, silent revenge donwvotes... Happy New Year!

Using the change of variables $x_1=a/b$, $x_2=b/c$, $x_3=c/a$, one asks for the maximum of $T$ under the constraint $S=0$, on the domain $D$ defined by $x_1\gt0$, $x_2\gt0$, $x_3\gt0$, where $$ T=\sum\limits_{k=1}^3R(x_k),\qquad R(x)=x/(3+x^2),\qquad S=x_1x_2x_3-1. $$ The extrema in $D$ are located at points such that the gradients of $T$ and $S$ are colinear. For every $k$, $\partial_kT=R'(x_k)$ with $R'(x)=(3-x^2)/(3+x^2)^2$ and $\partial_kS=(x_1x_2x_3)/x_k=1/x_k$, hence the condition is that $U(x_k)=x_kR'(x_k)$ does not depend on $k$.

The function $x\mapsto U(x)$ is smooth on $x\geqslant0$, increasing-then-decreasing and nonnegative on $0\leqslant x\leqslant\sqrt3$, and decreasing and negative on $x\gt\sqrt3$. Assume that $U(x_1)=U(x_2)=U(x_3)$ and call $v$ their common value. If $v\lt0$, the equation $U(x)=v$ has only one solution $x_v\gt1$ hence $x_1=x_2=x_3=x_v$ and $S\ne0$, which is absurd. If $v\gt0$, the equation $U(x)=v$ has at most two solutions in $(0,\sqrt3)$ hence either $x_1=x_2=x_3$, then their common value is $1$, or the $x_k$ are not all equal, then two of them are equal to some $x$ and the third one to $1/x^2$ and $U(x)=U(1/x^2)$. This last condition reads $W(x)=0$ with $$ W(x)=(3-x^2)(1+3x^4)^2-x(3x^4-1)(x^2+3)^2, $$ which has no solution $x\geqslant0$ except $x=1$. Finally, the gradients of $T$ and $S$ are colinear at the point $(1,1,1)$ and only there hence the only extremum on $D$ is $T(1,1,1)=3/4$, which is a local maximum since, for example, $T(x,1/x,1)\to1/4\lt3/4$ when $x\to0^+$.

To see what happens on the boundary of $D$, introduce the interval $K=[1/2,6]$. Then $R(x)\leqslant2/13$ for every $x$ not in $K$ and $R(x)\leqslant1/(2\sqrt3)$ for every $x\gt0$. Hence, as soon as one coordinate $x_k$ is not in $K$, $T\leqslant2/13+2\cdot1/(2\sqrt3)=0.731$. Since $0.731\lt3/4$, this proves that the supremum of $T$ is reached in $K\times K\times K$, and finally that this supremum is the maximum $T(1,1,1)=3/4$.

Caveat: The assertions above about the variations of the function $U$ and the roots of the polynomial $W$ were checked by inspecting W|A-drawn (parts of the) graphs of these two functions. To complete the proof, one should show them rigorously.