Inequalities

If a,ba,b and cc are sides of a triangle ABCABC with area XX, prove that ab+bc+ca43Xab + bc + ca \geq 4 \sqrt3 X.

#Algebra

Note by Subham Subian
4 years, 8 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

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 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

By rearrangement inequality, a2+b2+c2ab+ac+bca^2 + b^2 + c^2 \geq ab + ac + bc, then apply Weitzenböck's inequality, the result follows.

Pi Han Goh - 4 years, 8 months ago

Log in to reply

ok but i have a doubt we know that a^2+b^2+c^2>=ab+bc+ca and a^2+b^2+c^2>=4(square root of 3)(X) but how u got the relation between these two since both are less than a^2+b^2+c^2

Subham Subian - 4 years, 8 months ago

Log in to reply

Are you sure you got your question right? After reading your question for the second time, I'm pretty sure your inequality sign should be reversed.

Pi Han Goh - 4 years, 8 months ago

Log in to reply

@Pi Han Goh it was in a book(challenge and thrills of pre college mathematics) but i write the question write

Subham Subian - 4 years, 8 months ago

@Pi Han Goh The correct question is to prove;

 a2+b2+c243X\large\ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }\ge 4\sqrt { 3 } X

Priyanshu Mishra - 4 years, 8 months ago

Log in to reply

@Priyanshu Mishra It is evident from Paul-Erdos Inequality

Priyanshu Mishra - 4 years, 8 months ago

Counterexample: Consider the 3-4-5 triangle for which we have (a,b,c)=(3,4,5)(a,b,c)=(3,4,5) and hence ab+bc+ca=12+20+15=47ab+bc+ca=12+20+15=47 whereas the area X=3×42=6X=\frac{3\times 4}2=6 from which we have 43X=24341.5694\sqrt 3X=24\sqrt 3\approx 41.569, so we have ab+bc+ca>43Xab+bc+ca\gt 4\sqrt 3X


However, when the inequality sign is reversed, the claim is correct and the proof for that is given by Pi Han Goh in his comment.

Prasun Biswas - 4 years, 8 months ago

Log in to reply

Thanks! I've reversed the sign.

Calvin Lin Staff - 4 years, 7 months ago
×

Problem Loading...

Note Loading...

Set Loading...