2011 USAMO Problems/Problem 1

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.

Solution 3

Given $a^2+b^2+c^2+(a+b+c)^2 \leq 4$ , prove that $\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ac+1}{(a+c)^2} \geq 3.$

From the given constraint $a^2+b^2+c^2+(a+b+c)^2 \leq 4$ , we can expand to obtain: $a^2+b^2+c^2+a^2+b^2+c^2+2ab+2bc+2ca \leq 4$ $2(a^2+b^2+c^2)+2(ab+bc+ca) \leq 4$ $\frac{a^2+b^2+c^2}{2}+\frac{ab+bc+ca}{2} \leq 1.$

Now, observe that we can rewrite the left-hand side of our inequality as: \begin{align*} &\sum_{\text{cyc}} \frac{ab+1}{(a+b)^2} \\ &= \sum_{\text{cyc}} \frac{ab+\frac{a^2+b^2+c^2}{2}+\frac{ab+bc+ca}{2}}{(a+b)^2} \\ &= \sum_{\text{cyc}} \left[\frac{ab+\frac{a^2+b^2}{2}}{(a+b)^2} + \frac{\frac{c^2}{2}+\frac{bc+ca}{2}}{(a+b)^2}\right] \\ &= \sum_{\text{cyc}} \left[\frac{ab+\frac{a^2+b^2}{2}}{(a+b)^2}\right] + \sum_{\text{cyc}} \left[\frac{\frac{a^2+c^2}{2}}{(a+c)^2} + \frac{\frac{ac+bc}{2}}{(a+b)^2}\right]. \end{align*}

We now evaluate each part separately. Note that $\frac{ab+\frac{a^2+b^2}{2}}{(a+b)^2} = \frac{\frac{(a+b)^2}{2}}{(a+b)^2} = \frac{1}{2}.$ Therefore, $\sum_{\text{cyc}} \frac{ab+\frac{a^2+b^2}{2}}{(a+b)^2} = \frac{3}{2}.$ By the RMS-AM inequality, we have $(a+c)^2 \leq 2(a^2+c^2)$ , which gives $\frac{1}{(a+c)^2} \geq \frac{1}{2(a^2+c^2)}$ . Thus, \begin{align*} \sum_{\text{cyc}} \frac{\frac{a^2+c^2}{2}}{(a+c)^2} &\geq \sum_{\text{cyc}} \frac{\frac{a^2+c^2}{2}}{2(a^2+c^2)} \\ &= \sum_{\text{cyc}} \frac{1}{4} = \frac{3}{4}. \end{align*} The remaining terms can be written as: \begin{align*} \sum_{\text{cyc}} \frac{\frac{ac+bc}{2}}{(a+b)^2} &= \frac{1}{2}\sum_{\text{cyc}} \frac{c(a+b)}{(a+b)^2} \\ &= \frac{1}{2}\sum_{\text{cyc}} \frac{c}{a+b} \\ &= \frac{1}{2}\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right). \end{align*} By Nesbitt's inequality, $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2},$ so this part contributes at least $\frac{3}{4}$ . Adding all three parts together: $\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ac+1}{(a+c)^2} \geq \frac{3}{2}+\frac{3}{4}+\frac{3}{4} = 3.$

Therefore, the inequality is proven. $\square$

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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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