# Show $|\sqrt{|x|}-\sqrt{|y|}|=\frac{| |x|-|y| |}{\sqrt{|x|}+\sqrt{|y|}}$. [closed]

In a proof in class we used that for any real $$x$$ and $$y$$ we have $$\left|\sqrt{|x|}-\sqrt{|y|}\right|=\frac{| |x|-|y| |}{\sqrt{|x|}+\sqrt{|y|}}.$$However I'm not quite sure how one would show this and some help/hints would be appreciated. Thank you for your time.

• you could just drop the absolute values, assuming $x,y \ge 0$ Apr 27 at 10:48
• As long as $(x,y)\neq(0,0)$. Apr 27 at 11:42
• You can, with some more effort, also show that $|\sqrt{|x|}-\sqrt{|y|}|\le\sqrt{|x-y|}$ for the uniform continuity close to zero. Apr 27 at 15:55

If $$(x,y)\in\mathbb R^2\setminus \left\{0\right\}$$, then you can write the equivalent statement as follows:

$$||x|-|y||=\left|\left(\sqrt {|x|}\right)-\left(\sqrt {|y|}\right)\right|\times \left|\left(\sqrt {|x|}\right)+\left(\sqrt {|y|}\right)\right|$$

Or

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

which is correct.

This uses the simple fact $$(a-b)(a+b)=a^2-b^2.$$

Hint: without loss of generality, $$|x|\ge |y|$$ (since the expression is symmetric if you switch $$x$$ and $$y$$).