Let $a = \frac{m^{m+1} + n^{n+1}}{m^m + n^n}$, where $m$ and $n$ are positive integers. Prove that $a^m + a^n \geq m^m + n^n$.
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Tags: inequalities, AMC, USA(J)MO, USAMO, inequalities unsolved, 1991
11.10.2005 19:15
In your post, I boldfaced the part that you stated confusing conditions
15.10.2005 10:59
The problem is solved in the book" Winning Solutions" by Lozansky & Rousseau Here is a hint based on their solution: write a as m[ 1 + (a-m)/m] and n[ 1+ (a-n)/n] This is Question 4 in USAMO 1991.You can find a further solution at the Kalva site.
15.10.2005 11:28
I have lots of exercices using Bernoulli's inequality.
17.08.2007 22:17
13.04.2010 05:22
calc rulz wrote:
Your proof is not quite right.
14.04.2010 01:26
see here http://www.artofproblemsolving.com/Forum/viewtopic.php?p=280201 and the same method can prove the generalisation for $ n_i\in\mathbb N \ \ \ (i=1,...,k)$: $ \sum a^{n_i}\ge\sum {n_i}^{n_i}$ where $ a=\frac{\sum {n_i}^{n_i+1}}{\sum {n_i}^{n_i}}.$
21.07.2012 11:36
15.04.2017 18:00
Just another solution via Induction: Let the given statement be $P(m,n)$, base case is trivial. Now assume that its true for all $i\le m$ and $j\le n$ for some $m,n\in \mathbb{N}$. Its easy to see that $P(m,n)\Rightarrow P(m+1,n+1)$. Its easy to see that $a^{m+1}+a^n<m^{m+1}+n^n$ and $a^{m}+a^{n+1}<m^{m}+n^{n+1}$ cant be simultaneously true, if they were then adding both of them will give a contradiction. Thus at least one of them is false. But by symmetry both will be false thus $a^{m+1}+a^n\ge m^{m+1}+n^n$ and $a^{m}+a^{n+1}\ge m^{m}+n^{n+1}$ and we conclude that $P(m,n)\Rightarrow P(m+1,n)$ and $P(m,n)\Rightarrow P(m,n+1)$. Thus from the principle of induction the result holds.