Let $a = \frac{m^{m+1} + n^{n+1}}{m^m + n^n}$, where $m$ and $n$ are positive integers. Prove that $a^m + a^n \geq m^m + n^n$.
Problem
Source:
Tags: inequalities, AMC, USA(J)MO, USAMO, inequalities unsolved, 1991
10000th User
11.10.2005 19:15
liangchene wrote:
Here is a problem u can solve with Bernoulli's inequality:
The theorem goes like this
for x>1 and x>-1 (or r an even integer and x any real)
\[ (1+x)^r \geq 1+xr \]
the question is :
Let \[ a = \frac{m^{m+1} + n ^ {n+1}}{m^m+n^n} \] for m,n positive intergers. Prove that $a^m + a^n \geq m^m+n^n$
I have no idea how to start this problem, can anybody give me a hint?
In your post, I boldfaced the part that you stated confusing conditions
Assume WLOG $m\ge n$ because the expression in $a$ and the inequality are symmetric.
Play with division: make binomials that look like this form: $1+f(m,n)$
Michael Niland
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.
Cezar Lupu
15.10.2005 11:28
I have lots of exercices using Bernoulli's inequality.
calc rulz
17.08.2007 22:17
Because $ m$, $ n$, and $ a$ are all positive, $ f(x) = x^{m}$ is convex between $ a$ and $ m$, so it lies above the tangent line through $ m$.
The equation of this line is $ L(x) = m^{m}+f'(m)(x-m) = m^{m}+m^{m}(x-m)$
Thus $ a^{m}\ge L(a) = m^{m}+m^{m}(a-m)$.
Similarly, $ a^{n}\ge n^{n}+n^{n}(a-n)$.
Adding these gives $ a^{m}+a^{n}\ge m^{m}+n^{n}+m^{m}(a-m)+n^{n}(a-n)$.
But $ m^{m}(a-m)+n^{n}(a-n) = (m^{m}+n^{n})(a)-m^{m+1}-n^{n+1}= 0$, so we have that $ a^{m}+a^{n}\ge m^{m}+n^{n}+0$ as desired.
QED
Differ
13.04.2010 05:22
calc rulz wrote:
Because $ m$, $ n$, and $ a$ are all positive, $ f(x) = x^{m}$ is convex between $ a$ and $ m$, so it lies above the tangent line through $ m$.
The equation of this line is $ L(x) = m^{m} + f'(m)(x - m) = m^{m} + m^{m}(x - m)$
Thus $ a^{m}\ge L(a) = m^{m} + m^{m}(a - m)$.
Similarly, $ a^{n}\ge n^{n} + n^{n}(a - n)$.
Adding these gives $ a^{m} + a^{n}\ge m^{m} + n^{n} + m^{m}(a - m) + n^{n}(a - n)$.
But $ m^{m}(a - m) + n^{n}(a - n) = (m^{m} + n^{n})(a) - m^{m + 1} - n^{n + 1} = 0$, so we have that $ a^{m} + a^{n}\ge m^{m} + n^{n} + 0$ as desired.
QED
Your proof is not quite right.
spanferkel
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}}.$
SBS1
21.07.2012 11:36
Note that \[a=m(1+\frac{a-m}{m})=n(1+\frac{a-n}{n})\]
Using the Bernoulli's inequality ,\[a^m=m^m(1+\frac{a-m}{m})^m \ge m^m(1+a-m)\] and \[a^n=m^m(1+\frac{a-n}{n})^n \ge n^n(1+a-n)\] Adding we have ${a^m+a^n\ge (m^m+n^n)+a(m^m+n^n)-(m^{m+1}}+n^{n+1})$.Putting the value of $a$,we get \[a^m+a^n\ge (m^m+n^n)\] as desired
$\blacksquare$
dovakin123
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.