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Please help to solve the following inequality using rearrangement inequalities.

Let $a \gt 0$, $b \gt0$ and $ab=1$. Prove that \begin{equation}\frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}.\end{equation}

Thanks.

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You have asked so far some 26 questions (in 2 months), from which about 14 (!!) are about inequalities. I think it is about time you show a littel effort, some ideas, background... –  DonAntonio Sep 5 '12 at 13:50
    
@DonAntonio I'm sure that the ideas will appear. I want to improve my knowledges about inequalities - I hope you haven't a problem with my wish. –  Iuli Sep 5 '12 at 13:54
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What DonAntonio means (I believe) is that you should probably at least mention some things you've tried in your questions. –  Cameron Buie Sep 5 '12 at 14:06

5 Answers 5

up vote 5 down vote accepted

We can assume $a \leq 1 \leq b$. Applying rearrangement inequalities to $$ \begin{align} a &\leq 1 \\ 1 &\leq b \end{align} $$ we get $$ a + b \geq 1 + ab = 2 $$ and $$ b + 3a \geq 2a + 2 \\ a + 3b \geq 2b + 2 $$ Therefore $$ \begin{align} \frac{a}{a^2+3}+\frac{b}{b^2+3} &= \frac{1}{a + 3b} + \frac{1}{b + 3a} \leq\\ &\frac{1}{2b + 2} + \frac{1}{2a + 2} = \frac{a}{2a + 2} + \frac{1}{2a + 2} = \frac 1 2 \end{align} $$

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+1 very nice solution ! –  vanna Sep 6 '12 at 19:23

I don't have a rearrangement inequality proof yet, but I really like the following proof I got.

First note that $a+b \ge 2 \sqrt{ab} = 2$ by AM-GM.

$a^2 + 3 = a^2 + 3ab = a(a+3b) \ge a(2 + 2b) = 2ab(a + 1) = 2(a+1)$, and $b^2 + 3 = b^2 + 3ab = b(b+3a) \ge b(2 + 2a) = 2b(1+a)$.

Thus, we have $$\frac{a}{a^2+3}+\frac{b}{b^2+3} \le \frac{a}{2(a+1)} + \frac{1}{2(a+1)} = \frac{1}{2}$$ and we're through!

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Very slick! ${}$ –  Cameron Buie Sep 5 '12 at 14:23
    
you can change $a*b=1$ as $a=1/b$ put into equation,you get cubic from which logically you can reduce it to quadratic and solve it –  dato datuashvili Sep 5 '12 at 14:32
    
@dato of course! But that proof wouldn't be too nice now, would it? –  Rijul Saini Sep 5 '12 at 14:36
    
@dato: Actually, you get a quartic--even less nice! (I originally took that approach and typed it up as answer, but then remembered that the OP wanted to use rearrangement inequalities, and so I deleted it.) –  Cameron Buie Sep 5 '12 at 17:10

Since $ab=1$, we have $b = \frac{1}{a}$, and thus your inequality is equivalent to $\displaystyle\frac{a}{a^2+3}+\frac{a}{1+3a^2}\leq\frac{1}{2}$.

You can simply define $f(a) = \displaystyle\frac{a}{a^2+3}+\frac{a}{1+3a^2}$ and maximize $f$.

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Where's Rearrangement Inequality involved? –  Rijul Saini Sep 5 '12 at 14:03
    
Yes, I know . But I want an answer using rearrangements inequalities. Thanks, anyway :) –  Iuli Sep 5 '12 at 14:04

Homogenize the given problem into, $$\frac{\sqrt{ab}}{a+3b}+\frac{\sqrt{ab}}{b+3a}\leq \frac 12.$$ Now note that, using the AM-GM inequality we have $a+3b\geq 2\sqrt{2b(a+b)},$ so that $$\dfrac{\sqrt{ab}}{a+3b}\leq \frac{\sqrt{a}}{2\sqrt{2(a+b)}}.$$ Hence it suffices to check that $\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{2(a+b)}}\leq 1,$ which is perfectly true from the Cauchy-Schwarz inequality. $\Box$

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An opportunity to show the defects of AM-GM, you get a weaker inequailty.

$\dfrac{a^2+3}{a}=a+\dfrac{3}{a} \ge 2 \sqrt{3} \implies (\dfrac{a^2+3}{a})^{-1} \le \dfrac{1}{2 \sqrt{3}}$

$\dfrac{b^2+3}{b}=b+\dfrac{3}{b} \ge 2 \sqrt{3} \implies (\dfrac{b^2+3}{b})^{-1} \le \dfrac{1}{2 \sqrt{3}}$

When you add them, you have weaker inequality. Therefore, re-arrangement inequality is an appropriate one!

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