# Rearranging $|\sqrt{x}-\sqrt{y}|$

I'm just going through some examples in my Analysis notes and we're looking at Holder continuity of $\sqrt{x}$ on $[0,1]$. One of the steps is

$$\left|\sqrt{x}-\sqrt{y} \right|=\frac{\left|x-y \right|}{\sqrt{x}+\sqrt{y}}$$

I've been fiddling around with the left hand side for half an hour and cannot figure out how that works. I know that you can define the absolute value as $|x| = \sqrt{x^2}$, but I can't get it to work.

Squares' difference: $\;a^2-b^2=(a-b)(a+b)\;$ , and now put $\;a=\sqrt x\;,\;\;b=\sqrt y\;$ .
$\left|\sqrt{x}-\sqrt{y} \right|= \left|\sqrt{x}-\sqrt{y} \right|\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}} = \frac{\left|x-y\right|}{\sqrt{x}+\sqrt{y}}$, as required.