Proof Contest Day 3

ab+1(a+b)2+bc+1(b+c)2+ca+1(c+a)2≥3\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)2≤4a^2+b^2+c^2+(a+b+c)^2 \le 4. Prove the above inequality.

#Algebra

Note by Department 8
5 years, 5 months ago

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Comments

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+γ−α2b=α+γ−β2c=α+β−γ2\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+γ2≤4,\alpha^2 + \beta^2 + \gamma^2 \le 4,

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

γ2−(α−β)2+44γ2+α2−(β−γ)2+44α2+β2−(γ−α)2+44β2≥γ2−(α−β)2+α2+β2+γ24γ2+α2−(β−γ)2+α2+β2+γ24α2+β2−(γ−α)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+γα2β2≥32.\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 - 5 years, 5 months ago

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Great solution!

Adarsh Kumar - 5 years, 3 months ago

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Thank you:-)

Department 8 - 5 years, 3 months ago

This is 2011 USAMO #1 I believe

Anand Iyer - 5 years, 5 months ago

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Yes it is.

Department 8 - 5 years, 5 months ago

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 - 5 years, 4 months ago
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