# Prove the inequality $\sum\limits_{cyc}\sqrt{x+yz}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}$

Let $x,y,z$ is positive real numbers, such that $\frac1x+\frac1y+\frac1z=1$. Prove the inequality $$\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}$$

### My work so far:

$\frac1x+\frac1y+\frac1z=1\Rightarrow xyz=xy+yz+zx$

$\sqrt{x+yz}=\sqrt{x+xyz-xz-xy}=\sqrt x \cdot \sqrt{(y-1)(z-1)}$

An homogenization gives $$\sum\limits_{cyc}\sqrt{x+yz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}\geq\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}$$ or $$\sum\limits_{cyc}\sqrt{yz(x+y)(x+z)}\geq xy+xz+yz+\sum\limits_{cyc}x\sqrt{yz},$$ which is C-S: $$\sqrt{yz(x+y)(x+z)}=\sqrt{(xy+yz)(zx+yz)}\geq x\sqrt{yz}+yz$$ Done!