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# Proof Contest Day 3

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

Let $$a,b$$ and $$c$$ be positive real numbers such that $$a^2+b^2+c^2+(a+b+c)^2 \le 4$$. Prove the above inequality.

Note by Lakshya Sinha
1 year, 4 months ago

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Here is my solution for the problem:

Since

\begin{align*} (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{align*}

Let

\begin{align*} \alpha &= b + c \\ \beta &= c + a \\ \gamma &= a + b \end{align*}

This implies

\begin{align*} a &= \frac{\beta + \gamma - \alpha}2 \\ b &= \frac{\alpha + \gamma - \beta}2 \\ c &= \frac{\alpha + \beta - \gamma}2 \end{align*}

With this change of variables, the constraint becomes

$\alpha^2 + \beta^2 + \gamma^2 \le 4,$

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

\begin{align*} & \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{align*}

Therefore it remains to prove that

$\frac{\alpha\beta}{2\gamma^2} + \frac{\beta\gamma}{2\alpha^2} + \frac{\gamma\alpha}{2\beta^2} \ge \frac32.$.

Which is obviously true by AM-GM. · 1 year, 4 months ago

Great solution! · 1 year, 2 months ago

Thank you:-) · 1 year, 2 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 · 1 year, 4 months ago

This is 2011 USAMO #1 I believe · 1 year, 4 months ago

Yes it is. · 1 year, 4 months ago

Comment deleted Jan 29, 2016

$$a+2b+c<4$$ and $$a+b+c < 3$$ does not imply that $$b<1$$. For example, $$a=0.1, b=1.5, c=0.1$$ fulfills the two inequalities but has $$b>1$$.

$$a+2b+c<4$$ and $$a+b+c>3$$, however, does imply $$b<1$$. · 1 year, 3 months ago

I am going to sleep right now. :( Sorry, will continue my solution tomorrow. (It may contain a few minor mistakes!) · 1 year, 4 months ago