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

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.

@@ 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>
-==Solution==
+==Solutions==
+===Solution 1===
 Since
 <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.
+===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]]

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

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

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