Showing that $ \frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}$ for $a,b > 0$ and $ab = 1$ using rearrangement inequalities

Question

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.

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 05 '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 05 '12 at 13:54
What DonAntonio means (I believe) is that you should probably at least mention some things you've tried in your questions. — Cameron Buie, Sep 05 '12 at 14:06

AlbertH · Accepted Answer · 2012-09-07T13:53:13.700

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 $$ \begin{align} \frac{a}{a^2+3}+\frac{b}{b^2+3} &= \frac{1}{a + 3b} + \frac{1}{b + 3a} \leq\\ &\frac{1}{2b + 2} + \frac{1}{2a + 2} = \frac{a}{2a + 2} + \frac{1}{2a + 2} = \frac 1 2 \end{align} $$

score 2 · Answer 2 · edited Sep 05 '12 at 14:22

2

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!

edited Sep 05 '12 at 14:22

Cameron Buie

102,994

answered Sep 05 '12 at 14:17

Rijul Saini

2,544

Very slick! ${}$ – Cameron Buie Sep 05 '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 05 '12 at 14:32
@dato of course! But that proof wouldn't be too nice now, would it? – Rijul Saini Sep 05 '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 05 '12 at 17:10

score 0 · Answer 3 · answered Sep 05 '12 at 14:03

0

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 $f$.

answered Sep 05 '12 at 14:03

S4M

850

Where's Rearrangement Inequality involved? – Rijul Saini Sep 05 '12 at 14:03
Yes, I know . But I want an answer using rearrangements inequalities. Thanks, anyway :) – Iuli Sep 05 '12 at 14:04

score 0 · Answer 4 · answered Apr 09 '13 at 08:27

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. $\Box$

score 0 · Answer 5 · answered Apr 09 '13 at 09:33

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!

Showing that $ \frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}$ for $a,b > 0$ and $ab = 1$ using rearrangement inequalities

5 Answers5