Difference between revisions of "1991 USAMO Problems/Problem 4"
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Latest revision as of 06:39, 19 July 2016
Problem
Let 
where 
and 
are positive integers. Prove that 
.
[You may wish to analyze the ratio 
for real 
and integer 
.]
Solution
Let us assume without loss of generality that 
. We then note that
![\[m-a = \frac{m^{m+1} + m \cdot n^n}{m^m+n^n} - \frac{m^{m+1} - n^{n+1}}{m^m+n^n} = n^n \frac{m-n}{m^m+n^n} \qquad (*) .\]](http://latex.artofproblemsolving.com/b/b/c/bbc03cc4b825f2ea67970af098201732b65bfceb.png)
Similarly,
![\[a-n = m^m \frac{m-n}{m^m+n^n} \qquad (**) .\]](http://latex.artofproblemsolving.com/d/d/6/dd61876435249a1d13c5525267c0b6567cb9eecb.png)
We note that equations 
and 
imply that 
. Then 
, so
![\[\frac{1}{m} \sum_{i=0}^{m-1} (a/m)^i \le 1 \le \frac{1}{n} \sum_{i=0}^{n-1} (a/n)^i .\]](http://latex.artofproblemsolving.com/3/f/e/3fe72fd45c4467a873d1477b1822ce697a038138.png)
Multiplying this inequality by 
, we have
![\[n^n \frac{(m-n)}{m^m+n^n} \sum_{i=0}^{m-1} a^i m^{m-1-i} \le m^m \frac{(m-n)}{m^m+n^n} \sum_{i=0}^{n-1} a^i n^{n-1-i} .\]](http://latex.artofproblemsolving.com/5/6/7/567be9f8c8c13a07610c834771c4ebec92eebd8a.png)
It then follows that
Rearranging this inequality, we find that 
, as desired. 
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
| 1991 USAMO (Problems • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
