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

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

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

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

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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? – Chaebum Sheen · 2 weeks, 1 day ago

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– Gurīdo Cuong · 2 weeks ago

I came by it while doing some trigonometry exercisesLog in to reply

I've solved it i guess but wanna check my solution – Akram Mohamed · 2 weeks, 1 day ago

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