gobathegreat wrote: Let $a$, $b$ and $c$ be positive real numbers such that $a+b+c=1$. Prove the inequality: $\frac{1}{\sqrt{(a+2b)(b+2a)}}+\frac{1}{\sqrt{(b+2c)(c+2b)}}+\frac{1}{\sqrt{(c+2a)(a+2c)}} \geq 3$ $$\frac{1}{\sqrt{(a+2b)(b+2a)}}+\frac{1}{\sqrt{(b+2c)(c+2b)}}+\frac{1}{\sqrt{(c+2a)(a+2c)}} \geq \sum_{cyc} \frac{2}{3(a+b)}\geq 3$$

Troll problem, just looks scary

P.S. Same solution as above, just a bit detailed

By AM-GM it's quite clear that: $$\sqrt{(a+2b)(b+2a)}\leq \frac{3(a+b)}{2} \implies \sum \frac{1}{\sqrt{(a+2b)(b+2a)}}\geq \sum \frac{2}{3(a+b)}$$Therefore it suffices to prove that: $$\sum \frac{2}{3(a+b)}=\frac{2}{3}(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})\geq3$$Which we can do by Titu's Lemma as follows: $$\frac{2}{3}(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})\geq\frac{2}{3}(\frac{9}{2(a+b+c)})=\frac{2}{3}\cdot\frac{9}{2}=3$$

Easy P3 Its just Titu and Cauchy