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

(Solution by change of variables and AM-GM)
 
m (Solution 2)
 
(9 intermediate revisions by 5 users not shown)
Line 3: Line 3:
 
<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>
  
==Solution==
+
==Solutions==
 +
===Solution 1===
 
Since
 
Since
 
<cmath>
 
<cmath>
Line 57: Line 58:
  
 
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.
 +
 +
===Solution 2===
 +
Rearranging the condition yields that
 +
<cmath>a^2 + b^2 + c^2 +ab+bc+ac \le 2</cmath>
 +
 +
Now note that
 +
<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>
 +
 +
Summing this for all pairs of <math>\{ a,b,c \}</math> gives that
 +
<cmath>\sum_{cyc} \frac{2ab+2}{(a+b)^2} \ge 3+ \sum_{cyc}\frac{(c+a)(c+b)}{(a+b)^2} \ge 6</cmath>
 +
 +
By AM-GM. Dividing by <math>2</math> gives the desired inequality.
 +
 +
{{MAA Notice}}
 +
 +
==See also==
 +
{{USAMO newbox|year=2011|beforetext=|before=First Problem|num-a=2}}
 +
{{USAJMO newbox|year=2011|num-b=1|num-a=3}}
 +
[[Category:Olympiad Algebra Problems]]
 +
[[Category:Olympiad Inequality Problems]]

Latest revision as of 13:57, 24 October 2020

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)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6
All USAJMO Problems and Solutions
Invalid username
Login to AoPS