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
1 year, 2 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} \( \boxed{123} \)

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 - 1 year, 2 months ago

Log in to reply

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 - 1 year, 1 month ago

Log in to reply

Comment deleted Sep 30, 2016

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 - 1 year, 1 month ago

Log in to reply

I came by it while doing some trigonometry exercises

Gurīdo Cuong - 1 year, 2 months ago

Log in to reply

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

Akram Mohamed - 1 year, 2 months ago

Log in to reply

i hate this kind of problem!

李 增鑫 - 1 year ago

Log in to reply

Why?

Chaebum Sheen - 1 year ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...