How can we prove the Inequality : $ \frac {n!}{ 2^{n-1}((\frac {n-1}{2})!)^2} \leq \sqrt{n}$ How can we prove the following inequality? For every odd positive integer $n$,
$$ \frac {n!}{ 2^{n-1}((\frac {n-1}{2})!)^2} \leq \sqrt{n}$$
Thank You.

For completeness, here is the proof by induction:
Base Case:  The case $ n = 1$ is easy to check.
Inductive Step:  Assume that $$\frac{k!}{2^{k-1} \left( \frac{k-1}{2} \right)!^2} \leq \sqrt{k}.$$  We aim to show that this implies $$\frac{(k+2)!}{2^{k+1} \left( \frac{k+1}{2} \right)!^2} \leq \sqrt{k+2}.$$
We have
\begin{align}
\frac{(k+2)!}{2^{k+1} \left( \frac{k+1}{2} \right)!^2} &= \frac{(k+2)(k+1)}{4(\frac{k+1}{2})^2} \cdot \frac{k!}{2^{k-1} \left( \frac{k-1}{2} \right)!^2}
\\
&\leq \frac{(k+2)(k+1)}{4(\frac{k+1}{2})^2} \cdot \sqrt{k}
\\
&= \frac{k+2}{k+1} \sqrt{k}
\\
&\leq \sqrt{k+2}
\end{align}
which occurs if and only if $\left( \frac{k+2}{k+1}\right)^2 k \leq k+2$.  I will leave the last inequality for you to check.
