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In a solution for a test, I came upon the following: we now use $|\sqrt{x}-\sqrt{y}| \le \sqrt{|x-y|}$ (prove).

I've been unable to solve this - I've looked at the proof of the triangle inequality, but I haven't been able to apply the same concepts here.

I'd appreciate any help.

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    $\begingroup$ It may be easier if you rewrite the statement with as few square roots as possible. I assume $x,y$ are non-negative, right? Let $x=t^2$, $y=s^2$. Then the left hand side is $|t-s|$ and the right hand side is $\sqrt{|t^2-s^2|}$. You can factor the expression in the right hand side, so you are left with $\sqrt{|t-s|}\le\sqrt{t+s}$, or $|t-s|\le t+s$, which should be clear enough. $\endgroup$ – Andrés E. Caicedo Dec 11 '12 at 19:59
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Assume $x\ge y\ge 0$. Then we can do away with the absolute value signs. With that we get, from the original inequality, that $$ \sqrt x - \sqrt y \leq \sqrt{x-y} $$ Squaring both sides, we get $$ x -2\sqrt{xy} + y \leq x-y \quad \Rightarrow \quad 2\sqrt{xy}\geq 2y $$ which follows immediately from the assumption that $x\geq y$

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  • $\begingroup$ So assuming $\sqrt{x} - \sqrt{y} \leq \sqrt{x - y}$ you prove that $2\sqrt{xy} \geq 2xy$. So? $\endgroup$ – WimC Dec 11 '12 at 20:03
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    $\begingroup$ Well, solving equations and inequalities are usually done "the wrong way", going from what we want to solve, and working towards something we know is right. The trick is recognizing which steps are reversable, and which ones are not. So yes, I assumed the inequality in question held, and I showed that $\sqrt{xy}\geq y$. But, reading backwards, you can see that if you assume $\sqrt{xy}\geq y$, you can work your way towards the top, and see that it does indeed work out. $\endgroup$ – Arthur Dec 11 '12 at 20:17
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HINT: look at the square the inequality, that is, prove:

$$|\sqrt x - \sqrt y|^2\leq|x-y|$$

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