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 $x,y \ge 0$ $\endgroup$– gt6989bApr 27 at 10:48
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$\begingroup$ As long as $(x,y)\neq(0,0)$. $\endgroup$– Michael HoppeApr 27 at 11:42
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$\begingroup$ You can, with some more effort, also show that $|\sqrt{|x|}-\sqrt{|y|}|\le\sqrt{|x-y|}$ for the uniform continuity close to zero. $\endgroup$– Lutz LehmannApr 27 at 15:55
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2 Answers
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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 $y$).
answered Apr 27 at 10:29