Waste less time on Facebook — follow Brilliant.
×

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
3 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

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 · 2 months, 3 weeks ago

Log in to reply

@Chaebum Sheen 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 · 2 months, 1 week ago

Log in to reply

Comment deleted 2 months ago

Log in to reply

@Chaebum Sheen 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. Chaebum Sheen · 2 months, 1 week ago

Log in to reply

@Chaebum Sheen I came by it while doing some trigonometry exercises Gurīdo Cuong · 2 months, 3 weeks ago

Log in to reply

I've solved it i guess but wanna check my solution Akram Mohamed · 2 months, 3 weeks ago

Log in to reply

@Akram Mohamed i hate this kind of problem! 李 增鑫 · 1 month, 1 week ago

Log in to reply

@李 增鑫 Why? Chaebum Sheen · 1 month, 1 week ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...