# Inequality with square roots : $\frac{\sqrt{b+c}}{a}+\frac{\sqrt{c+a}}{b}+\frac{\sqrt{a+b}}{c}\geq \frac{4(a+b+c)}{\sqrt{(a+b)(b+c)(c+a)}}$

Let $a, b, c$ be positive real numbers. Prove that $\frac{\sqrt{b+c}}{a}+\frac{\sqrt{c+a}}{b}+\frac{\sqrt{a+b}}{c}\geq \frac{4(a+b+c)}{\sqrt{(a+b)(b+c)(c+a)}}$

Please check if my work is correct or not.

$\displaystyle\sum_{cyc}\frac{\sqrt{b+c}}{a}\geq \frac{4(a+b+c)}{\sqrt{(a+b)(b+c)(c+a)}}$

It's sufficient to show that $\displaystyle\sum_{cyc}\frac{(b+c)\sqrt{(a+b)(c+a)}}{a}\geq 4(a+b+c)$

By C-S,

$\displaystyle\sum_{cyc}\frac{(b+c)\sqrt{(a+b)(c+a)}}{a}\geq\displaystyle\sum_{cyc}\frac{(b+c)(a+\sqrt{bc})}{a}$

$RHS = \displaystyle\sum_{cyc}(b+c) + \sum_{cyc}\frac{(b+c)\sqrt{bc}}{a} = \displaystyle\sum_{cyc}(b+c) + \sum_{cyc}\frac{2\sqrt{bc}\sqrt{bc}}{a} = 2\sum_{cyc}a + \sum_{cyc}\frac{bc}{a} = 4\sum_{cyc}a$

We're done.

• @Macavity, typo was edited. Thanks. Jan 10, 2017 at 12:26
• Is my work correct ? Jan 11, 2017 at 1:59
• I think your methos It's right! Jan 11, 2017 at 2:45
• Thank you very much, math110. Jan 11, 2017 at 13:49