Inequality for Gamma functions Let $k, n ,m \in N$ and such that $0\leq k \leq n \leq m$. When the following ineuality is true?
$$
\frac{2^{m-k}\Gamma(n+1)\Gamma\left(\left[\frac{m+1-k}{2}\right]\right)\Gamma(m+1-n)}{\Gamma(m+1)\Gamma\left(\left[\frac{m+2-k}{2}\right]\right)}\leq  \sqrt{\pi}
$$
Thank you for your help.
 A: I'm assuming that "$[\cdots]$" is the floor function, sometimes written "$\lfloor \cdots \rfloor$".
Since $m$ and $n$ are integers we have
$$\Gamma(n+1) = n! \qquad \Gamma(m+1) = m! \qquad \Gamma(m+1-n) = (m-n)!$$
This allows us to rewrite the left-hand side of the inequality as
$$
2^{m-k} \frac{\Gamma\left(\left[\frac{m+1-k}{2}\right]\right)}{\Gamma\left(\left[\frac{m+2-k}{2}\right]\right)} \cdot \frac{n!(m-n)!}{m!} = 2^{m-k} \frac{\Gamma\left(\left[\frac{m+1-k}{2}\right]\right)}{\Gamma\left(\left[\frac{m+2-k}{2}\right]\right)} \cdot \frac{1}{\binom{m}{n}}.
$$
Now if $m-k$ is even then $m+2-k$ is even, which implies that $(m+2-k)/2$ is an integer and
$$
\left[\frac{m+2-k}{2}\right] = \frac{m+2-k}{2} = \frac{m-k}{2} + 1.
$$
This tells us that
$$
\Gamma\left(\left[\frac{m+2-k}{2}\right]\right) = \Gamma\left(\frac{m-k}{2} + 1\right) = \frac{m-k}{2} \,\Gamma\left(\frac{m-k}{2}\right).
$$
Then since
$$
\frac{m+1-k}{2} = \frac{m+2-k}{2} - \frac{1}{2},
$$
we have
$$
\left[\frac{m+1-k}{2}\right] = \frac{m+2-k}{2} - 1 = \frac{m-k}{2},
$$
which gives us
$$
\Gamma\left(\left[\frac{m+1-k}{2}\right]\right) = \Gamma\left(\frac{m-k}{2}\right).
$$
Some cancellation happens in the inequality and we are left with
$$
\frac{2^{m+1-k}}{(m-k)\binom{m}{n}} \leq \sqrt{\pi}.
$$
After multiplying by the denominator and taking logarithms this is equivalent to

$$
k \geq m+1 - \log_2\left[\sqrt{\pi}(m-k)\binom{m}{n}\right] \qquad \text{if } m-k \text{ is even.}
$$

If $m-k$ is odd then similar considerations reveal that
$$
\Gamma\left(\left[\frac{m+1-k}{2}\right]\right) = \Gamma\left(\left[\frac{m+2-k}{2}\right]\right),
$$
which leaves the inequality in the form
$$
\frac{2^{m-k}}{\binom{m}{n}} \leq \sqrt{\pi}.
$$
By again multiplying by the denominator and taking logarithms we see that this is equivalent to

$$
k \geq m - \log_2\left[\sqrt{\pi}\binom{m}{n}\right] \qquad \text{if } m-k \text{ is odd.}
$$

