# 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 .

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No suggestions? –  Frank Jul 7 '12 at 19:55

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.

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

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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 $a=b=c$

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