If $a, b, c\in\mathbb R^+,$ then prove that: $\sum_{cyc} \frac1{c-b}\left(\frac1{\sqrt{a+2b}}-\frac1{\sqrt{a+2c}}\right)\ge\frac3{\sqrt{(a+b+c)^3}}.$

Question

We have:$$\sum_{cyc}\frac1{c-b}\left(\frac1{\sqrt{a+2b}}-\frac1{\sqrt{a+2c}}\right)\\\\=\sum_{cyc}\frac{\sqrt{a+2c}-\sqrt{a+2b}}{(c-b)\sqrt{a+2b}\sqrt{a+2c}}\\\\=\sum_{cyc}\frac{2(c-b)}{(c-b)\sqrt{a+2b}\sqrt{a+2c}\left(\sqrt{a+2c}+\sqrt{a+2b}\right)}\\\\=\underbrace{\sum_{cyc}\frac2{(a+2c)\sqrt{a+2b}+(a+2b)\sqrt{a+2c}}}_{=E\text{ (say)}}.$$ So we need to show that: $E\ge\frac3{\sqrt{(a+b+c)^3}}.$

I tried AM $\ge$ HM and obtained: $$E\ge\frac{18}{\sum_{cyc}\left\{(a+2c)\sqrt{a+2b}+(a+2b)\sqrt{a+2c}\right\}}.$$ But now I'm confused. How to tackle it further!?

Please suggest me what to do next.. Thanks in advance.

Answer 1

Now, by AM-GM and Jensen we obtain: $$\sum_{cyc}\frac{2}{\sqrt{(a+2b)(a+2c)}(\sqrt{a+2b}+\sqrt{a+2c})}\geq$$ $$\geq\sum_{cyc}\frac{2}{\frac{a+2b+a+2c}{2}\cdot\sqrt{2(a+2b+a+2c)}}=\frac{3}{\sqrt{(a+b+c)^3}}.$$