# 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?

• Your last two inequalities are not true. Apply Gautschi's inequality with $x = \frac{{m - 1}}{2}$ and $s = \frac{1}{2}$ instead.
– Gary
May 16, 2022 at 12:59
• @Gary You should make that an answer. May 16, 2022 at 14:08

Gautschi's inequality states that $$x^{1 - s} < \frac{{\Gamma (x + 1)}}{{\Gamma (x + s)}}$$ for $$x>0$$ and $$0. Let $$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)}}.$$
This follows from the convexity of the gamma function (which is easily proved by twice differentiating the definition under the integral sign): $$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$$.