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$n\geq m\geq 3$, 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}2012-04-05

2 Answers 2

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$$(n+1)^m=n^m\Bigl(1+{1\over n}\Bigr)^m

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    (-1) This is not a solution, since you used unproved inequality $(1+1/n)2012-04-05
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    @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 $[n+1]^m\to[m]\times[n]^m$.2012-04-05
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    I think OP asked for the proof from the first principles. It is better to ask OP, until then (-1)2012-04-05
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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.