Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1991 USAMO Problems/Problem 4"
(→‎Problem)
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
  
−
Let <math>\, a =(m^{m+1} + n^{n+1})/(m^m + n^n), \,</math> where <math>\,m\,</math> and <math>\,n\,</math> are positive integers.  Prove that <math>\,a^m + a^n \geq m^m + n^n</math>.
+
Let <math>\, a =\frac{m^{m+1} + n^{n+1}}{m^m + n^n}, \,</math> where <math>\,m\,</math> and <math>\,n\,</math> are positive integers.  Prove that <math>\,a^m + a^n \geq m^m + n^n</math>.
  
−
[You may wish to analyze the ratio <math>\,(a^N - N^N)/(a-N),</math> for real <math>\, a \geq 0 \,</math> and integer <math>\, N \geq 1</math>.]
+
[You may wish to analyze the ratio <math>\,\frac{a^N - N^N}{a-N},</math> for real <math>\, a \geq 0 \,</math> and integer <math>\, N \ge 1</math>.]
  
 
== Solution ==
 
== Solution ==
Line 27: Line 27:
 
{{alternate solutions}}
 
{{alternate solutions}}
  
−
== Resources ==
+
== See Also ==
  
 
{{USAMO box|year=1991|num-b=3|num-a=5}}
 
{{USAMO box|year=1991|num-b=3|num-a=5}}

Latest revision as of 20:23, 21 November 2024

Problem

Let $\, a =\frac{m^{m+1} + n^{n+1}}{m^m + n^n}, \,$ 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 $\,\frac{a^N - N^N}{a-N},$ for real $\, a \geq 0 \,$ and integer $\, N \ge 1$.]

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.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_USAMO_Problems/Problem_4&oldid=235255"