If $x+y+z=3$, where $x$, $y$ and $z$ are sides of a triangle, prove that $$\frac{x}{\sqrt y}+\frac{y}{\sqrt z}+\frac{z}{\sqrt x} \geq 3$$
We have $0< x,y,z<\frac{3}{2}$ from the triangle inequality.
Here's my attempt:
Multiply both sides by $\sqrt {xyz}$: $$x \sqrt{xz}+y \sqrt{xy}+z \sqrt{xz} \geq3 \sqrt{xyz}$$ I am stuck here because if I use AM-GM on the left side it wouldn't work. I also tried to square both sides but it didn't help.