Proof that $|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|},\quad x,y \geq 0$ Any hints on how I can prove the inequality:

$$|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|},\quad x,y \geq 0$$

Thank you.

Case 1:  If $x = y$, both sides equal zero.  
Case 2:  If $x \neq y$, both sides of the inequality are positive, so we can square both sides of 
$$|\sqrt{x} - \sqrt{y}| \leq \sqrt{|x - y|}$$
to obtain
$$|\sqrt{x} - \sqrt{y}|^2 \leq |x - y| = |(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})| = |\sqrt{x} + \sqrt{y}||\sqrt{x} - \sqrt{y}|$$
Divide both sides of the inequality by $|\sqrt{x} - \sqrt{y}|$ to obtain
$$|\sqrt{x} - \sqrt{y}| \leq |\sqrt{x} + \sqrt{y}|$$
Since both sides of the inequality are positive, we can square both sides to obtain
$$x - 2\sqrt{xy} + y \leq x + 2\sqrt{xy} + y$$
Cancelling $x + y$ yields
$$-2\sqrt{xy} \leq 2\sqrt{xy}$$
Dividing both sides of the inequality by $-2\sqrt{xy}$ yields
$$0 \geq -1$$
Since the steps are reversible, the original inequality holds.
