$n\geq m\geq 3$, prove $m\cdot n^m>(n+1)^m$.
I will use mathematical induction to prove, would like to know another proof
$n\geq m\geq 3$, prove $m\cdot n^m>(n+1)^m$.
I will use mathematical induction to prove, would like to know another proof
$(n+1)^m=n^m\Bigl(1+{1\over n}\Bigr)^m
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.