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

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 · 10 months, 1 week ago

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How did you come up with those points?

Thanks for contributing and helping other members aspire to be like you! – Calvin Lin Staff · 9 months, 3 weeks ago

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– Chaebum Sheen · 9 months, 3 weeks ago

The only time that I had seen such an inequality that there were two terms on the right but only one term on the left was generally related to the triangle inequality. So, I tried to find these points, and saw that \[\frac{|x-y|}{\sqrt{x^2+1}\sqrt{y^2+1}} \times \prod_{cyc} \sqrt{x^2+1} = \sqrt{(zx-zy)^2+(x-y)^2} \] However, this is just a handwavy explanation as it was just an idea that I had.Log in to reply

– Gurīdo Cuong · 10 months, 1 week 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 · 10 months, 1 week ago

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– 李 增鑫 · 8 months, 3 weeks ago

i hate this kind of problem!Log in to reply

– Chaebum Sheen · 8 months, 3 weeks ago

Why?Log in to reply