I would like to prove, that for $m \ge 2$ we have the following inequality:
$$\frac{\sqrt 2\ \Gamma\left(\frac m 2\right)}{\sqrt{m - 1}\ \Gamma\left(\frac{m - 1}{2}\right)} < 1$$
My thoughts
This inequality is equivalent to that one:
$$\sqrt 2\ \Gamma\left(\frac m 2\right) < \sqrt{m - 1}\ \Gamma\left(\frac{m - 1}{2}\right)$$
And now, when $m \ge 2$, then of course $\sqrt{m - 1} \ge 2$, but I don't know how can I prove, that
$$\Gamma\left(\frac m 2\right) < \Gamma\left(\frac{m - 1}{2}\right)$$
For example, when $m \ge 2$ and $m = 2k$, where $k \in \mathbb N$. We have that
$$(k - 1)! < \Gamma\left(k - \frac 1 2\right)$$
But how this inequality can be justified?