Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$n\geq m\geq 3$, prove $m\cdot n^m>(n+1)^m$.

I will use mathematical induction to prove, would like to know another proof

share|cite|improve this question
Rewrite it as $1+\frac{1}{n}< m^{\frac{1}{m}}$. Then you just need to show that $1+\frac{1}{m}<m^{\frac{1}{m}}$ – Thomas Andrews Apr 5 '12 at 14:34
up vote 5 down vote accepted

$$(n+1)^m=n^m\Bigl(1+{1\over n}\Bigr)^m<n^m\bigl(e^{1/n}\bigr)^m\leq e\ n^m<m\ n^m$$

share|cite|improve this answer
(-1) This is not a solution, since you used unproved inequality $(1+1/n)<e^{1/n}$ – Norbert Apr 5 '12 at 14:45
@Norbert: $1+x\leq e^x$ is a standard calculus fact. The OP did ask for a "non inductive proof"; he didn't ask for an injection $[n+1]^m\to[m]\times[n]^m$. – Christian Blatter Apr 5 '12 at 15:31
I think OP asked for the proof from the first principles. It is better to ask OP, until then (-1) – Norbert Apr 5 '12 at 17:11

This condition is equivalent to the condition that $m > ({1 + 1/n})^m$. As $n \geq m$, the following condition implies this:

$m > ({1+1/m})^m$.

In other words,

$m^{m+1} > ({m+1})^m$

or $m^{1/m} > ({m+1})^{1/({m+1})}$

Consider the function $f(x) = x^{1/x}$.

Consider $g(x) = log(f(x)) = \frac{1}{x} log(x)$

$g'(x) = \frac{1}{x^2}(1-log(x))$, which is $0$ only at $x = e$. We can easily check this is a maximum (by differentiating again, etc). So for $m \geq 3$, as $m \geq e$, $g(m) > g(m+1)$. So $f(m) > f(m+1)$. This proves the claim.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.