1991 USAMO Problems/Problem 4

Problem

Let $\, a =(m^{m+1} + n^{n+1})/(m^m + n^n), \,$ where $\,m\,$ and $\,n\,$ are positive integers. Prove that $\,a^m + a^n \geq m^m + n^n$ .

[You may wish to analyze the ratio $\,(a^N - N^N)/(a-N),$ for real $\, a \geq 0 \,$ and integer $\, N \geq 1$ .]

Solution

Let us assume without loss of generality that $m\ge n$ . We then note that $m-a = \frac{m^{m+1} + m \cdot n^n}{m^m+n^n} - \frac{m^{m+1} - n^{n+1}}{m^m+n^n} = n^n \frac{m-n}{m^m+n^n} \qquad (*) .$ Similarly, $a-n = m^m \frac{m-n}{m^m+n^n} \qquad (**) .$

We note that equations $(*)$ and $(**)$ imply that $n \le a \le m$ . Then $a/m \le 1 \le a/n$ , so $\frac{1}{m} \sum_{i=0}^{m-1} (a/m)^i \le 1 \le \frac{1}{n} \sum_{i=0}^{n-1} (a/n)^i .$ Multiplying this inequality by $m^m n^n(m-n)/(m^m+n^n)$ , we have $n^n \frac{(m-n)}{m^m+n^n} \sum_{i=0}^{m-1} a^i m^{m-1-i} \le m^m \frac{(m-n)}{m^m+n^n} \sum_{i=0}^{n-1} a^i n^{n-1-i} .$ It then follows that $\begin{align*} m^m - a^m &= (m-a) \sum_{i=0}^{m-1} a^i m^{m-1-i} = n^n \frac{m-n}{m^m+n^n} \sum_{i=0}^{m-1} a^i m^{m-1-i} \\ &\le m^m \frac{m-n}{m^m+n^n} \sum_{i=0}^{n-1} a^i n^{n-1-i} = (a-n) \sum_{i=0}^{n-1} a^i n^{n-1-i} \\ &= a^n - n^n . \end{align*}$ Rearranging this inequality, we find that $a^m + a^n \ge m^m + n^n$ , as desired. $\blacksquare$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

1991 USAMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

Discussion on AoPS/MathLinks

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1991 USAMO Problems/Problem 4

Problem

Solution

Resources