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.
 A: 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 $x>\frac{n+1}{2}$.]
A: Showing the first is equivalent to showing that
$$n^{n/2} \leq n!$$
We may approach this by taking logarithms and comparing to an integral, i.e.,
\begin{align}
\ln(n!) &= \sum_{k=1}^n \ln(k)\\
&\geq \int_1^n \ln(x)\;dx
= \left[x \ln(x) - x\right]_1^n\\
&= n\left(\ln(n)-1\right) + 1\\
&\geq n \frac{\ln(n)}{2}
\end{align}
For $n \geq e^2$, which follows because
$$ \left(\ln(n) -1 \right) - \frac{\ln(n)}{2} = \tfrac12 \ln(n) - 1$$
which is increasing and has a root at $e^2$.
For $n < e^2 < 8$ you could just show this from direct computation.
It follows that:
$$\ln(n!) \geq \frac{n}{2}\ln(n) = \ln(n^{n/2})$$
and by the monotonicity of $\ln(x)$,
$$n^{n/2} \leq n! \implies \sqrt{n} \leq \sqrt[n]{n!}$$
