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I want to prove the following inequality: $$\frac{\sqrt a+\sqrt b+\sqrt c}{2}\ge\frac{1}{\sqrt a}+\frac{1}{\sqrt b}+\frac{1}{\sqrt c}$$ Where $a,b,c$ are positive reals and with the horrible condition: $$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2$$ It was a problem of the MEMO 2011. What I tried so far is the following:

Make the substitution $x=\frac{a}{1+a}$, $y=\frac{b}{1+b}$, $z=\frac{c}{1+c}$ so $a=\frac{x}{1-x}$, $b=\frac{y}{1-y}$, $c=\frac{z}{1-z}$. The conditions then change to $x+y+z=2$ and $0<x,y,z<1$ since $\frac{a}{1+a}<1$ and so the other two terms. The original inequality changes to: $$\frac{\sqrt \frac{x}{1-x}+\sqrt \frac{y}{1-y}+\sqrt \frac{z}{1-z}}{2}\ge\sqrt \frac{1-x}{x}+\sqrt \frac{1-y}{y}+\sqrt \frac{1-z}{z}$$ Now we have $y+z=2-x>1>x$ and similarly $x+y>z$, $z+x>y$. This is equivalent to $x=u+v$, $y=v+w$ and $z=w+u$. So with this change of variables, the condition $0<x,y,z<1$ disappears. The other condition becomes $(u+v)+(v+w)+(w+u)=2 \iff u+v+w=1$. The inequality changes to: $$\frac{\sqrt \frac{u+v}{1-u-v}+\sqrt \frac{v+w}{1-v-w}+\sqrt \frac{w+u}{1-w-u}}{2}\ge\sqrt \frac{1-u-v}{u+v}+\sqrt \frac{1-v-w}{v+w}+\sqrt \frac{1-w-u}{w+u} \iff \frac{\sqrt \frac{u+v}{w}+\sqrt \frac{v+w}{u}+\sqrt \frac{w+u}{v}}{2}\ge\sqrt \frac{w}{u+v}+\sqrt \frac{u}{v+w}+\sqrt \frac{v}{w+u} \iff \sqrt \frac{u+v}{2w}+\sqrt \frac{v+w}{2u}+\sqrt \frac{w+u}{2v}\ge\sqrt \frac{2w}{u+v}+\sqrt \frac{2u}{v+w}+\sqrt \frac{2v}{w+u}$$ Am I right so far? The last inequality seems kind of "beauty" since each term on the righthand side has its reciprocal on the lefthand side. But I wasn't able to find a solution from there, so any help is highly appreciated! Thanks.

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The condition $\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2 \iff abc = a+b+c+2$, so as you deduced through two successive substitutions, we can have $$a = \frac{u+v}w, \, b = \frac{v+w}u, \, c = \frac{w+u}v$$ to get $$\sum_{cyc} \sqrt{\frac{u+v}w} \ge 2 \sum_{cyc} \sqrt{\frac{w}{u+v}} \iff \sum_{cyc} \frac{u+v-2w}{\sqrt{w(u+v)}}\ge 0$$

Now note that $u+v-2w$ and $w(u+v)$ are arranged oppositely and apply Chebyshev's inequality. Use $\sum_{cyc} (u+v-2w)=0$ to conclude.

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  • $\begingroup$ Yes, from here it is easy to conclude. Thank you, a very nice solution! $\endgroup$ – Redundant Aunt Sep 11 '14 at 4:53
  • $\begingroup$ You're welcome... $\endgroup$ – Macavity Sep 11 '14 at 5:02
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Proceeding from your last line: by the power mean inequality (or more simply the AM-HM actually, which is a special case) it is sufficient that $(\sqrt\frac{2w}{u+v}+\sqrt \frac{2u}{v+w}+\sqrt \frac{2v}{w+u})^2 \leq 9$ (left to you! it's not very hard). Motivation: the power mean inequality can give us an inequality between some numbers and their reciprocals, which is just what we were looking for.

Hence we require $\sqrt\frac{2w}{u+v}+\sqrt \frac{2u}{v+w}+\sqrt \frac{2v}{w+u} \leq 3$. By Cauchy-Schwarz, all we then need is

$\frac{2w}{u+v}+\frac{2u}{v+w}+\frac{2v}{w+u} \leq 3$, or

$\frac{2-2(u+v)}{u+v}+\frac{2-2(v+w)}{v+w}+\frac{2-2(w+u)}{w+u} \leq 3$, or

$\frac{1}{u+v}+\frac{1}{v+w}+\frac{1}{w+u} \geq \frac{3}{2}$.

$f(x)=\frac{1}{x}$ is convex, so by Jensen's inequality $\frac{1}{u+v}+\frac{1}{v+w}+\frac{1}{w+u} \geq 3 \frac{3}{(u+v)+(v+w)+(w+u)}=\frac{3}{2}$, and we're done!

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  • $\begingroup$ Actually, I've thought about it, it's all fine. Ask me to elaborate tomorrow if my (albeit small) leap with the AM-HM is a bit much. Hope this answers your question nicely/adequately. $\endgroup$ – Shakespeare Sep 10 '14 at 22:11
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    $\begingroup$ Excuse me if I'm wrong, but aren't you mistaking with the equivalence $\frac{2-2(u+v)}{u+v}+\frac{2-2(v+w)}{v+w}+\frac{2-2(w+u)}{w+u}\le 3$ and $\frac{1}{u+v}+\frac{1}{v+w}+\frac{1}{w+u}\ge \frac{3}{2}$? Shouldn't it be equivalent to $\frac{1}{u+v}+\frac{1}{v+w}+\frac{1}{w+u}\le \frac{9}{2}$? $\endgroup$ – Redundant Aunt Sep 10 '14 at 22:41
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    $\begingroup$ By Nesbitt's inequality, $\frac{2w}{u+v}+\frac{2u}{v+w}+\frac{2v}{w+u} \ge 3$. $\endgroup$ – Macavity Sep 11 '14 at 1:43
  • $\begingroup$ Ok, it was written sufficiently late at night. Sorry about that! $\endgroup$ – Shakespeare Sep 11 '14 at 6:50

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