[dektep]

Am-Gm ; $x^2+yz \ge 2\sqrt{x^2yz} = 2\sqrt{x}$
$\therefore x^2+y^2+z^2+xy+yz+zx \ge 2(\sqrt{x}+\sqrt{y}+\sqrt{z})$
$\longrightarrow \frac{x^2+y^2+z^2+xy+yz+zx}{\sqrt{x} + \sqrt{y} + \sqrt{z}}\geq 2$