Difference between revisions of "2011 USAMO Problems/Problem 1"

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[[Category:Olympiad Inequality Problems]]

Revision as of 21:23, 16 April 2016

Problem

Let $a$, $b$, $c$ be positive real numbers such that $a^2 + b^2 + c^2 + (a + b + c)^2 \le 4$. Prove that \[\frac{ab + 1}{(a + b)^2} + \frac{bc + 1}{(b + c)^2} + \frac{ca + 1}{(c + a)^2} \ge 3.\]

Solutions

Solution 1

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*} it is natural to consider a change of variables: \begin{align*} \alpha &= b + c \\ \beta &= c + a \\ \gamma &= a + b \end{align*} with the inverse mapping given by: \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 \[\frac{\alpha\beta}{2\gamma^2} + \frac{\beta\gamma}{2\alpha^2} + \frac{\gamma\alpha}{2\beta^2} \ge \frac32.\]

We note that the product of the three (positive) terms is 1/8, therefore by AM-GM their mean is at least 1/2, and thus their sum is at least 3/2 and we are done.

Solution 2

Rearranging the condition yields that \[a^2 + b^2 + c^2 +ab+bc+ac \le 2\]

Now note that \[\frac{2ab+2}{(a+b)^2} \ge \frac{2ab+a^2 + b^2 + c^2 +ab+bc+ac}{(a+b)^2}=\frac{(a+b)^2 + (c+a)(c+b)}{(a+b)^2}\]

Summing this for all pairs of $\{ a,b,c \}$ gives that \[\sum_{cyc} \frac{2ab+2}{(a+b)^2} \ge 3+ \sum_{cyc}\frac{(c+a)(c+b)}{(a+b)^2} \ge 6\]

By AM-GM. Dividing by $2$ gives the desired inequality. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

See also

2011 USAMO (ProblemsResources)
First Problem Followed by
Problem 2
1 2 3 4 5 6
All USAMO Problems and Solutions
2011 USAJMO (ProblemsResources)
1 Followed by
Problem 3
1 2 3 4 5 6
All USAJMO Problems and Solutions
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