Proving $\sqrt{n} \le \sqrt[n]{n!}$

Question

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.

Answer 1

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 $x>\frac{n+1}{2}$.]

Answer 2

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 $\ln(x)$, $$n^{n/2} \leq n! \implies \sqrt{n} \leq \sqrt[n]{n!}$$