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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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    What Do$n$A$n$tonio $m$ea$n$s (I believe) is that you should probably at least $m$ention some things you've tried in your questions.2012-09-05

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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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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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    @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.)2012-09-05
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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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    Yes, I know . But I want an answer using rearrangements inequalities. Thanks, anyway :)2012-09-05
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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!