Show $|\sqrt{|x|}-\sqrt{|y|}|=\frac{| |x|-|y| |}{\sqrt{|x|}+\sqrt{|y|}}$.

Question

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.

Answer 1

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 $$(a-b)(a+b)=a^2-b^2.$$

Answer 2

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