$n^{n+1} \geq (n+1)^{n} , n \in \mathbb{N}, n > 2$ Problematic induction proof, I need to prove that:
$$n^{n+1} \geq (n+1)^{n} , n \in \mathbb{N}, n > 2$$
I mean it is pretty obvious that it is true, but I can not prove that with induction.
for $n + 1$ I get that:
$$(n+1)^{n+2} \geq (n+2)^{n+1}$$
$$(n+1)^{n}*(n+1)^2 \geq (n+2)^{n}*(n+2)$$
And I don't see how could I plug my thesis into the inequality. I could do something like this, but it doesn't get me anywhere.
$$(n+1)^{n}*(n+1)^2 \geq n^{n}*(n+1)^2 \geq  (n+2)^{n}*(n+2)$$
The problem here is $(n+2)^{n}$, I can make it only bigger to keep my inequality valid, even if I use Bernoulis inequality.
 A: Spoiler: this is not a hint, it is a solution.
The key is in the hint by @peanut . We can rewrite $n^{n+1} \ge (n+1)^n$ as:
$$ \frac{n^{n+1}}{n^n} \ge \frac{(n+1)^n}{n^n} $$
Or
$$ n \ge (1+ \frac{1}{n})^n $$
The base case is true: $3 > (1+\frac{1}{3})^3$ $\ $ (it is actually easier to see that $3^4 > 4^3$).
Now, if the inequality holds for up to $n$, we show that it also holds for $n+1$ .
Noting that $ \frac{1}{n} > \frac{1}{n+1} $ we can write:
$$ n \ge (1+ \frac{1}{n+1})^n $$
And
$$ n(1+ \frac{1}{n+1}) \ge (1+ \frac{1}{n+1})^{n+1} $$
Finally, because $ \frac {n}{n+1} < 1$ we have
$$ n + 1 \ge (1+ \frac{1}{n+1})^{n+1} $$
A: Consider the ratio
$r(n)
=\dfrac{(n+1)^n}{n^{n+1}}
$.
If we can show that
$r(n) \gt r(n+1)$
for all $n$ and
$r(n) < 1$
for some $n=n_0$,
then
$r(n) \lt 1$ for
$n \ge n_0$
so
$(n+1)^n < n^{n+1}
$
for $n \ge n_0$.
$r(3)
=\dfrac{4^3}{3^4}
=\dfrac{64}{81}
\lt 1
$
so we can take
$n_0 = 3$.
$\begin{array}\\
\dfrac{r(n+1)}{r(n)}
&=\dfrac{\dfrac{(n+2)^{n+1}}{(n+1)^{n+2}}}{\dfrac{(n+1)^n}{n^{n+1}}}\\
&=\dfrac{n^{n+1}(n+2)^{n+1}}{(n+1)^{n+2}(n+1)^n}\\
&=\dfrac{(n(n+2))^{n+1}}{(n+1)^{2n+2}}\\
&=\left(\dfrac{n(n+2)}{(n+1)^2}\right)^{n+1}\\
&=\left(\dfrac{n^2+2n}{n^2+2n+1}\right)^{n+1}\\
&\lt 1\\
\end{array}
$
and we are done.
A: If you take logs of both sides, what you need to show is
$${\ln( n + 1) \over n + 1 } \leq {\ln n \over n}$$
This follows from the fact that ${\displaystyle {\ln x \over x}}$ is decreasing for $x > e$.
