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Difference between revisions of "1991 USAMO Problems/Problem 4"

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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=344949#344949 Discussion on AoPS/MathLinks]
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[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]

Revision as of 20:52, 3 July 2013

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

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