\[\frac{|x-y|}{\sqrt{x^2+1}\sqrt{y^2+1}}+\frac{|y-z|}{\sqrt{y^2+1}\sqrt{z^2+1}}\geq\frac{|z-x|}{\sqrt{z^2+1}\sqrt{x^2+1}}\]

Let \(x,y,z\in\mathbb{R}\), prove this inequality above.

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## Comments

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TopNewestLet points \(A,B,C\) be such that \(A (xy,z), B(yz,x) C(zx,y)\). Note that from the triangle inequality we have that \[ \overline{AB} \le \overline{AC}+\overline{BC}\] From the Pythagorean theorem, it follows that \[|z-x|\sqrt{y^2+1} \le |x-y|\sqrt{z^2+1}+|y-z|\sqrt{x^2+1}\] Dividing both sides by \[\prod_{cyc} \sqrt{x^2+1}\] gives us \[\frac{|z-x|}{\sqrt{z^2+1}\sqrt{x^2+1}} \le \frac{|x-y|}{\sqrt{x^2+1}\sqrt{y^2+1}}+\frac{|y-z|}{\sqrt{y^2+1}\sqrt{z^2+1}}\] As desired.

This is my first time seeing such an inequality. Where did you come by it?

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Your solution is a great read. Nice usage of the geometric interpretation!

How did you come up with those points?

Thanks for contributing and helping other members aspire to be like you!

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Comment deleted Sep 30, 2016

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I came by it while doing some trigonometry exercises

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I've solved it i guess but wanna check my solution

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i hate this kind of problem!

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Why?

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