Proving $\frac{\sqrt 2\ \Gamma\left(\frac m 2\right)}{\sqrt{m - 1}\ \Gamma\left(\frac{m - 1}{2}\right)} < 1$ 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?

Gautschi's inequality states that
$$
x^{1 - s}  < \frac{{\Gamma (x + 1)}}{{\Gamma (x + s)}}
$$
for $x>0$ and $0<s<1$. Let $m>1$. Taking $x=\frac{m-1}{2}$ and $s=\frac{1}{2}$ gives
$$
\sqrt {\frac{{m - 1}}{2}}  < \frac{{\Gamma \left( {\frac{{m - 1}}{2} + 1} \right)}}{{\Gamma \left( {\frac{m}{2}} \right)}},
$$
i.e.,
$$
\frac{{\sqrt {\frac{{m - 1}}{2}} \Gamma \left( {\frac{m}{2}} \right)}}{{\Gamma \left( {\frac{{m - 1}}{2} + 1} \right)}} < 1.
$$
But note that
$$
\frac{{\sqrt {\frac{{m - 1}}{2}} \Gamma \left( {\frac{m}{2}} \right)}}{{\Gamma \left( {\frac{{m - 1}}{2} + 1} \right)}} = \frac{{\sqrt {\frac{{m - 1}}{2}} \Gamma \left( {\frac{m}{2}} \right)}}{{\frac{{m - 1}}{2}\Gamma \left( {\frac{{m - 1}}{2}} \right)}} = \frac{{\Gamma \left( {\frac{m}{2}} \right)}}{{\sqrt {\frac{{m - 1}}{2}} \Gamma \left( {\frac{{m - 1}}{2}} \right)}} = \frac{{\sqrt 2 \Gamma \left( {\frac{m}{2}} \right)}}{{\sqrt {m - 1} \Gamma \left( {\frac{{m - 1}}{2}} \right)}}.
$$
