Subadditivity of square root function Any hint to prove 
$$ \sqrt{x+y} \le \sqrt{x} + \sqrt{y}, \qquad \forall x,y \ge 0 $$

Another:
$x,y \geq 0$ then the product $\sqrt{x} \sqrt{y}$ have sence and now
$$\sqrt{x + y} \leq \sqrt{x + 2\sqrt{x}\sqrt{y} + y} = \sqrt{(\sqrt{x} + \sqrt{y})^2} = \sqrt{x} + \sqrt{y}$$
