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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.

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  • $\begingroup$ you could just drop the absolute values, assuming $x,y \ge 0$ $\endgroup$
    – gt6989b
    Apr 27 at 10:48
  • $\begingroup$ As long as $(x,y)\neq(0,0)$. $\endgroup$ Apr 27 at 11:42
  • $\begingroup$ You can, with some more effort, also show that $|\sqrt{|x|}-\sqrt{|y|}|\le\sqrt{|x-y|}$ for the uniform continuity close to zero. $\endgroup$ Apr 27 at 15:55

2 Answers 2

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

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

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