$x + y + z \geq \sqrt{xy} + \sqrt{yz} + \sqrt{zx}$
$\Leftrightarrow 2x + 2y + 2z \geq 2\sqrt{xy} + 2\sqrt{yz} + 2\sqrt{zx}$
$\Leftrightarrow 2x + 2y + 2z - 2\sqrt{xy} - 2\sqrt{yz} - 2\sqrt{zx} \geq 0$
$\Leftrightarrow \left ( x - 2\sqrt{xy} + y \right ) + \left ( y - 2\sqrt{yz} + z \right ) + \left ( z - 2\sqrt{zx} + x \right ) \geq 0$
$\Leftrightarrow \left ( x - y \right )^{2} + \left ( y - z \right )^{2} + \left ( z - x \right )^{2} \geq 0$ luôn đúng với mọi $x, y, z$
Dấu "=" xảy ra khi $x = y = z$
