How to show using proof by induction: $\sqrt[n]{n!} \leqslant \frac{n+1}{2}, n \in \mathbb{Z}^+$ I'm having quite a few problems with the following proof by induction question:
$$\sqrt[n]{n!} \leqslant \frac{n+1}{2}, n \in \mathbb{Z}^+$$
I manage to do the easy parts of the base step ($n=1$) and the hypothesis ($n=k$) but after that I don't know how to work with the equation and show that it is true for $n=k+1$.
Any help is greatly appreciated.
 A: If you want to do it without induction, by the way, it's immediate from the AM/GM inequality.
A: Hint: The challenge in induction problems with an inequality is the inequality.  You won't get the same formula on both sides like you would in an induction with an equality (you do some manipulation and the formulas pop out).
In this problem, you are correct that the $n=1$ case is fairly easy, so you assume that for $n=k$, the statement is true; i.e., $\sqrt[k]{k!}\leq\frac{k+1}{2}$.  Now, raise both sides to the $k^{\text{th}}$ power to get $k!\leq\left(\frac{k+1}{2}\right)^k$.  Next, multiply both sides by $(k+1)$ to get 
$
(k+1)!\leq(k+1)\cdot\left(\frac{k+1}{2}\right)^k=\frac{(k+1)^{k+1}}{2^k}.
$
Now, take the $k+1^{\text{th}}$ root of both sides to get that
$$
\sqrt[k+1]{(k+1)!}\leq\frac{k+1}{2^{k/(k+1)}}.
$$
Now, if you can show that
$$
\frac{k+1}{2^{k/(k+1)}}\leq\frac{k+2}{2},
$$
then you will be done.  This is equivalent to proving
$$
\sqrt[k+1]{2}\leq1+\frac{1}{k+1}.
$$
In induction-based inequalities, you often need to introduce this middle object and prove some different inequality about it.
