Proof Contest Day 3

ab+1(a+b)2+bc+1(b+c)2+ca+1(c+a)23\large{\frac { ab+1 }{ { ( a+b ) }^{ 2 } } +\frac { bc+1 }{ ( b+c ) ^{ 2 } } +\frac { ca+1 }{ { ( c+a ) }^{ 2 } } \ge 3}

Let a,ba,b and cc be positive real numbers such that a2+b2+c2+(a+b+c)24a^2+b^2+c^2+(a+b+c)^2 \le 4. Prove the above inequality.

Note by Department 8
3 years, 7 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   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 </span>...<span></span> ... <span> or </span>...<span></span> ... <span> 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

Sort by:

Top Newest

Here is my solution for the problem:

Since

(a+b)2+(b+c)2+(c+a)2=2(a2+b2+c2+ab+bc+ca)=a2+b2+c2+(a+b+c)2, \begin{aligned} (a+b)^2 + (b+c)^2 + (c+a)^2 &= 2(a^2 + b^2 + c^2 + ab + bc + ca) \\ &= a^2 + b^2 + c^2 + (a + b + c)^2, \end{aligned}

Let

α=b+cβ=c+aγ=a+b \begin{aligned} \alpha &= b + c \\ \beta &= c + a \\ \gamma &= a + b \end{aligned}

This implies

a=β+γα2b=α+γβ2c=α+βγ2\begin{aligned} a &= \frac{\beta + \gamma - \alpha}2 \\ b &= \frac{\alpha + \gamma - \beta}2 \\ c &= \frac{\alpha + \beta - \gamma}2 \end{aligned}

With this change of variables, the constraint becomes

α2+β2+γ24,\alpha^2 + \beta^2 + \gamma^2 \le 4,

while the left side of the inequality we need to prove is now

γ2(αβ)2+44γ2+α2(βγ)2+44α2+β2(γα)2+44β2γ2(αβ)2+α2+β2+γ24γ2+α2(βγ)2+α2+β2+γ24α2+β2(γα)2+α2+β2+γ24β2=2γ2+2αβ4γ2+2α2+2βγ4α2+2β2+2γα4β2=32+αβ2γ2+βγ2α2+γα2β2.\begin{aligned} & \frac{\gamma^2 - (\alpha - \beta)^2 + 4}{4\gamma^2} + \frac{\alpha^2 - (\beta - \gamma)^2 + 4}{4\alpha^2} + \frac{\beta^2 - (\gamma - \alpha)^2 + 4}{4\beta^2} \ge \\ & \frac{\gamma^2 - (\alpha - \beta)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\gamma^2} + \frac{\alpha^2 - (\beta - \gamma)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\alpha^2} + \frac{\beta^2 - (\gamma - \alpha)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\beta^2} = \\ & \frac{2\gamma^2 + 2\alpha\beta}{4\gamma^2} + \frac{2\alpha^2 + 2\beta\gamma}{4\alpha^2} + \frac{2\beta^2 + 2\gamma\alpha}{4\beta^2} = \\ & \frac32 + \frac{\alpha\beta}{2\gamma^2} + \frac{\beta\gamma}{2\alpha^2} + \frac{\gamma\alpha}{2\beta^2}. \end{aligned}

Therefore it remains to prove that

αβ2γ2+βγ2α2+γα2β232.\frac{\alpha\beta}{2\gamma^2} + \frac{\beta\gamma}{2\alpha^2} + \frac{\gamma\alpha}{2\beta^2} \ge \frac32..

Which is obviously true by AM-GM.

Department 8 - 3 years, 7 months ago

Log in to reply

Great solution!

Adarsh Kumar - 3 years, 5 months ago

Log in to reply

Thank you:-)

Department 8 - 3 years, 5 months ago

Log in to reply

This is 2011 USAMO #1 I believe

Anand Iyer - 3 years, 7 months ago

Log in to reply

Yes it is.

Department 8 - 3 years, 7 months ago

Log in to reply

Given constraint: a^2+b^2+c^2+ab+bc+ca\leq2 Now, frac{(ab+1)/(a+b)^2}\geq1/2+1/2{(a+b)(b+c)}/(a+b)^2 Cyclic summing over three variables a,b,c And using A.M-G.M we obtain required result

kushal padole - 3 years, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...