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

Question

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?

Answer 1

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 $$ \frac{{\sqrt {\frac{{m - 1}}{2}} \Gamma \left( {\frac{m}{2}} \right)}}{{\Gamma \left( {\frac{{m - 1}}{2} + 1} \right)}} = \frac{{\sqrt {\frac{{m - 1}}{2}} \Gamma \left( {\frac{m}{2}} \right)}}{{\frac{{m - 1}}{2}\Gamma \left( {\frac{{m - 1}}{2}} \right)}} = \frac{{\Gamma \left( {\frac{m}{2}} \right)}}{{\sqrt {\frac{{m - 1}}{2}} \Gamma \left( {\frac{{m - 1}}{2}} \right)}} = \frac{{\sqrt 2 \Gamma \left( {\frac{m}{2}} \right)}}{{\sqrt {m - 1} \Gamma \left( {\frac{{m - 1}}{2}} \right)}}. $$

Answer 2

This follows from the convexity of the gamma function (which is easily proved by twice differentiating the definition under the integral sign): $$2 \Gamma \left(x+\frac12 \right) < \Gamma(x)+\Gamma(x+1) = \Gamma(x) \, (1+x)$$

Set $x=k-\frac12$:

$$2 \Gamma(k) < \Gamma \left(k-\frac12 \right) \, (k+\frac12)$$

$$ (k-1)! \frac{2}{k+\frac12} < \Gamma \left (k-\frac12 \right) $$

and your result follows for $k \geq 2$.