You may've known about this before

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

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

#Algebra

Easy Math Editor

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Math	Appears as
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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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`\boxed{123}`	$\boxed{123}$

Comments

Let 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 - 4 years, 9 months ago

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!

Calvin Lin Staff - 4 years, 8 months ago

I came by it while doing some trigonometry exercises

P C - 4 years, 9 months ago

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

Akram Mohamed - 4 years, 9 months ago

i hate this kind of problem!

李增鑫 - 4 years, 7 months ago

Why?

Chaebum Sheen - 4 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$