# Square root of absolute difference is greater than absolute of square root difference

For $$x,y \ge 0, |\sqrt{x} - \sqrt{y}| \le \sqrt{|x - y|}$$.

How to prove this.

I have tried by squaring both sides, but failed.

Can you help me out.

Hint:

• This is equivalent to showing $$|a-b| \le \sqrt{|a^2-b^2|}$$ for $$a,b \ge 0$$ where $$a=\sqrt{x}$$ and $$b=\sqrt{y}$$
• $$|a^2-b^2|=|(a-b)(a+b)| =|a-b| \, |a+b| \ge |a-b|^2$$ for $$a,b \ge 0$$ since $$|a-b| \le |a+b|$$, with equality when $$a=0$$ or $$b=0$$ or $$a=b$$

So you could write $$|\sqrt{x} - \sqrt{y}| = \sqrt{\left|\sqrt{x} - \sqrt{y}\right|\, \left|\sqrt{x} - \sqrt{y}\right|}\le \sqrt{\left|\sqrt{x} - \sqrt{y}\right|\, \left|\sqrt{x} + \sqrt{y}\right|} = \sqrt{\left|x - y\right|}$$

with equality when $$x=0$$ or $$y=0$$ or $$x=y$$

or if you square everything

$$\left(|\sqrt{x} - \sqrt{y}|\right)^2 = {\left|\sqrt{x} - \sqrt{y}\right|\, \left|\sqrt{x} - \sqrt{y}\right|}\le {\left|\sqrt{x} - \sqrt{y}\right|\, \left|\sqrt{x} + \sqrt{y}\right|} ={\left|x - y\right|}$$

Sketch: It comes down to comparing the squares, i.e. $$x+y-2\sqrt{xy}\le |x-y|.$$ We may suppose $$x\ge y\;$$ w.l.o.g., so it is equivalent to $$x+y-2\sqrt{xy}\le x-y\iff2y=2\sqrt{y^2}\le 2\sqrt{xy}.$$