# Prove two inequalities (sums of square roots of quotients of positive numbers)

Let $a,b,c$ be positive numbers. Prove that:

1. $$\sqrt{\frac{a}{a+2b+3c}}+\sqrt{\frac{b}{b+2c+3a}}+\sqrt{\frac{c}{c+2a+3b}} \leq \frac{\sqrt{6}}{2}$$

2. $$\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}} \leq 3$$

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–  Golbez Oct 18 '12 at 15:43
the first one do you have the solution of this? –  LevanDokite Oct 19 '12 at 5:04

1.By Cauchy's inequality \begin{align}LHS^2\leq \left(\sum_{cyc}c+2a+3b\right)\left(\sum_{cyc}\frac{a}{(a+2b+3c)(c+2a+3b)}\right)\end{align}

What remains to prove is \begin{align}\sum_{cyc}\frac{a}{(a+2b+3c)(c+2a+3b)}\leq \frac{1}{4(a+b+c)}\end{align}

which is equivalent to proving \begin{align}6\sum_{cyc} a^3+\sum_{cyc}a^2 b\geq\sum_{cyc}ab^2+18abc \end{align} which is obvious by AM-GM

2.By pigeon hole theorem two of $\sqrt{\frac{2a}{a+b}},\sqrt{\frac{2b}{b+c}},\sqrt{\frac{2c}{c+a}}$ must be greater or smaller than $1$, by symmetry let us assume that \begin{align}\left(\sqrt{\frac{2a}{a+b}}-1\right)\left(\sqrt{\frac{2b}{b+c}}-1\right)\geq 0\end{align} That is \begin{align} \sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}\leq 1+\sqrt{\frac{4ab}{(a+b)(b+c)}} \end{align} Also, By Cauchy's Inequality $\sqrt{(a+b)(b+c)}\geq\sqrt{ab}+\sqrt{bc}$ Thus \begin{align} \sqrt{\frac{4ab}{(a+b)(b+c)}}\leq \frac{2\sqrt{ab}}{\sqrt{ab}+\sqrt{bc}}=\frac{2\sqrt{a}}{\sqrt{a}+\sqrt{c}} \end{align}

Also, note that \begin{align} \sqrt{\frac{2c}{c+a}}=2\sqrt{\frac{c}{2(c+a)}}\leq \frac{2\sqrt{c}}{\sqrt{c}+\sqrt{a}} \end{align}

By the inequalities above \begin{align} \sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}}\leq 1+\sqrt{\frac{4ab}{(a+b)(b+c)}}+\frac{2\sqrt{c}}{\sqrt{c}+\sqrt{a}}\leq 3 \end{align}

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The second inequality is false! For $a = 16$ and $b = c = 1$ we have $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \geq \sqrt{\frac {2a} {b + c}} = \sqrt{\frac {2\cdot 16} {1 + 1}} = 4 > 3$$

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sorry I have checked :D and now it 's correct –  LevanDokite Oct 19 '12 at 5:04