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

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<cmath>\frac{ab + 1}{(a + b)^2} + \frac{bc + 1}{(b + c)^2} + \frac{ca + 1}{(c + a)^2} \ge 3.</cmath>
 
<cmath>\frac{ab + 1}{(a + b)^2} + \frac{bc + 1}{(b + c)^2} + \frac{ca + 1}{(c + a)^2} \ge 3.</cmath>
  
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==Solution==
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==Solution 1==
 
Since
 
Since
 
<cmath>
 
<cmath>
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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.
 
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.
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 +
==Solution 2==
 +
Rearranging the condition yields that
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<cmath>a^2 + b^2 + c^2 +ab+bc+ac \le 2</cmath>
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 +
Now note that
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<cmath>\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}</cmath>
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 +
Summing this for all pairs of <math>\{ a,b,c \}</math> gives that
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<cmath>\sum_{cyc} \frac{2ab+2}{(a+b)^2} \ge 3+ \sum_{cyc}\frac{(c+a)(c+b)}{(a+b)^2} \ge 6</cmath>
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 +
By AM-GM. Dividing by <math>2</math> gives the desired inequality.
  
 
==See also==
 
==See also==
 
{{USAMO newbox|year=2011|beforetext=|before=First Problem|num-a=2}}
 
{{USAMO newbox|year=2011|beforetext=|before=First Problem|num-a=2}}

Revision as of 18:22, 29 April 2011

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 \[\frac{ab + 1}{(a + b)^2} + \frac{bc + 1}{(b + c)^2} + \frac{ca + 1}{(c + a)^2} \ge 3.\]

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.

See also

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