Waste less time on Facebook — follow Brilliant.
×

An interesting inequality

\[\large \frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{1}{3}\geq\frac{8}{9}\bigg(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\bigg)\] Let \(a,b,c>0\), prove the above inequality.


This problem was taken from a discussion on VMF (Viet Maths Forum)

Note by Gurīdo Cuong
1 year, 5 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

Here's the solution. First, without losing generality, we can assume that \(ab+bc+ac=3\), the problem'll become \[a^2+b^2+c^2+1\geq\frac{8}{3}\bigg(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\bigg)\] Now we'll do some manipulation on the variables in \(LHS\) \[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a(a+b)(a+c)+b(b+a)(b+c)+c(c+a)(c+b)}{(a+b)(b+c)(c+a)}\] \[=\frac{a^3+b^3+c^3+(a+b+c)(ab+bc+ca)}{(a+b+c)(ab+bc+ca)-abc}=\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca+ab+bc+ca)+3abc}{3(a+b+c)-abc}\] \[=\frac{(a+b+c)(a^2+b^2+c^2)+3abc}{3(a+b+c)-abc}\] So now the inequality is \[a^2+b^2+c^2+1\geq\frac{8[(a+b+c)(a^2+b^2+c^2)+3abc]}{9(a+b+c)-3abc}\] \[\Rightarrow [9(a+b+c)-3abc](a^2+b^2+c^2+1)\geq 8(a+b+c)(a^2+b^2+c^2)+24abc\] \[\Leftrightarrow 9(a+b+c)(a^2+b^2+c^2)+9(a+b+c)-3abc(a^2+b^2+c^2)-3abc\geq 8(a+b+c)(a^2+b^2+c^2)+24abc\] \[\Leftrightarrow (a+b+c)(a^2+b^2+c^2+9)\geq 3abc(a^2+b^2+c^2+9)\] \[\Rightarrow a+b+c\geq 3abc\] The last inequality can be proven easily by these following inequalities: \((a+b+c)^2\geq 3(ab+bc+ca)\) and \(abc\leq\sqrt{\big(\frac{ab+bc+ca}{3}\big)^3}\). The equality holds when \(a=b=c\)

Gurīdo Cuong - 1 year, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...