<s>[INST] Prove that for any $x\geq0$ and $y \ge 0$ we have: $|\sqrt{x} − \sqrt{y}| \le \sqrt{|x − y|}.$ 
Prove that for any $x\geq0$ and $y \ge 0$ we have:   $$|\sqrt{x} − \sqrt{y}|   \le \sqrt{|x − y|}.$$

$$|\sqrt(x) - \sqrt(y)|^2 \le \sqrt{|x-y|}^2$$
$$(\sqrt{x}-\sqrt{y})^2 \le |x-y|$$
$$x - 2\sqrt{xy} + y \le |x-y|$$
How do I take the absolute value separately to prove it?
 [/INST] Consider $2$ cases:
$$\begin{align} & x<y \Rightarrow \sqrt{x}-\sqrt{y}<0< \sqrt{|x-y|}, \\
& x\ge y \Rightarrow \sqrt{x}-\sqrt{y}\le \sqrt{x-y} \Rightarrow x-2\sqrt{xy}+y\le x-y \Rightarrow y\le x. \end{align}$$
</s>