Inequality : $ \sum_{cyc} \frac{a+b}{\sqrt{a+2c}} \geq 2 \sqrt{a+b+c}$ 
Let $a$, $b$ and $c$ be positive real numbers. Prove that: 
  $$ \displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}} \geq 2 \sqrt{a+b+c}.$$

My attempt :
By Holder inequality,
$ \left(\displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}}\right)^2 \displaystyle\sum_{cyc}(a+2c)\geq \left(\displaystyle\sum_{cyc} \sqrt{a+b}\right)^3$
$ \left(\displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}}\right)^2 \geq \frac{\left(\displaystyle\sum_{cyc} \sqrt{a+b}\right)^3}{\displaystyle\sum_{cyc}(a+2c)}$
$\Leftrightarrow \left(\displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}}\right)^2 \geq \frac{\left(\displaystyle\sum_{cyc} \sqrt{a+b}\right)^3}{3\displaystyle\sum_{cyc}a}$
$\Leftrightarrow \displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}} \geq \sqrt{\frac{\left(\displaystyle\sum_{cyc} \sqrt{a+b}\right)^3}{3\displaystyle\sum_{cyc}a}}$

Your idea to use Holder was very good!
By Holder $$\left(\sum_{cyc}\frac{a+b}{\sqrt{a+2c}}\right)^2\sum_{cyc}(a+b)(a+2c)\geq8(a+b+c)^3.$$
Thus, it's enough to prove that
$$8(a+b+c)^3\geq4(a+b+c)\sum_{cyc}(a+b)(a+2c)$$ or
$$2(a+b+c)^2\geq\sum_{cyc}(a+b)(a+2c),$$ which is
$$\sum_{cyc}(a-b)^2\geq0.$$
Done!
