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

Question

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.

Answer 1

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

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

Answer 2

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 $$x+y-2\sqrt{xy}\le x-y\iff2y=2\sqrt{y^2}\le 2\sqrt{xy}.$$