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# 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.

Note by Gurīdo Cuong
1 year ago

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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? · 1 year ago

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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! Staff · 11 months, 3 weeks ago

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Comment deleted 11 months ago

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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. · 11 months, 3 weeks ago

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I came by it while doing some trigonometry exercises · 1 year ago

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I've solved it i guess but wanna check my solution · 1 year ago

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i hate this kind of problem! · 10 months, 3 weeks ago

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Why? · 10 months, 3 weeks ago

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