Proving $\sqrt{n} \le \sqrt[n]{n!}$ I need to prove one thing:
$$
n \ge 1:
\sqrt{n} \le \sqrt[n]{n!} \le \frac{n + 1}{2}
$$
The second part:
$$
\sqrt[n]{n!} \le \frac{n + 1}{2}
$$
is easy to proof.But the first is more complicated. Help please.

Hint: Show that, for integers $k$ such that $1\leq k\leq n$, $k(n+1-k)$ has a minimum when $k=1$ or $k=n$.
So: $$n!\cdot n!=(1\cdot n)\cdot(2\cdot (n-1))\cdots((n-1)\cdot 2)\cdots(n\cdot 1)\geq n^{n}$$
[Essentially, $f(x)=x(n+1-x)$ is increasing for $x<\frac{n+1}{2}$ and decreasing for $x>\frac{n+1}{2}$.]
