# Inequality $abcd=1\Rightarrow\sum\limits_{cyc}\frac{a}{b\sqrt{a+b+cd}}\geq\frac{4}{\sqrt3}$

Let $a$, $b$, $c$ and $d$ be positive numbers such that $abcd=1$. Prove that: $$\frac{a}{b\sqrt{a+b+cd}}+\frac{b}{c\sqrt{b+c+da}}+\frac{c}{d\sqrt{c+d+ab}}+\frac{d}{a\sqrt{d+a+bc}}\geq\frac{4}{\sqrt3}$$

I tried C-S, AM-GM, Holder and more, but without success.

Thank you!