Non-induction proof that for $n\geq m\geq 3$, $m\cdot n^m>(n+1)^m$

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

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

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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

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

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(-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.

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