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1991 USAMO Problems/Problem 4

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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} ... 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)
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 Followed by
Problem 5
All USAMO Problems and Solutions
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