Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 .

share|improve this question
    
No suggestions? –  Frank Jul 7 '12 at 19:55

3 Answers 3

up vote 4 down vote accepted

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.

share|improve this answer

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

share|improve this answer

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 $a=b=c$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.