Is there a way to mathematically show that
$$ \frac{1}{q} > 2^q \left( \frac{\Gamma(q)\Gamma(q+1)}{\Gamma(2q+1)} \right)$$
for all $1 < q < \infty$? This is true by plotting this on Mathematica/Desmos, but I can't seem to find an analytic proof for this. I have also noticed that $\left( \frac{\Gamma(q)\Gamma(q+1)}{\Gamma(2q+1)} \right)$ is nothing but $\text{Beta}(q,q+1)$ but I can't seem to find any relevant facts that I can use to prove this. Any help would be appreciated. Thanks!
On
Since $$2^q\,\frac{ \Gamma (q)\, \Gamma (q+1)}{\Gamma (2 q+1)}=2^{-q}\sqrt{\pi } \frac{ \Gamma (q)}{\Gamma \left(q+\frac{1}{2}\right)}$$ I suppose that you can easily show it using induction (the equality is for $q=1$).
If you want to use algebra, consider the function $$f(q)=\log \left(\frac{1}{q}\right)-\log\Bigg[2^{-q}\sqrt{\pi } \frac{ \Gamma (q)}{\Gamma \left(q+\frac{1}{2}\right)} \Bigg]$$ for which $$f'(q)=-\frac{1}{q}-\psi (q)+\psi \left(q+\frac{1}{2}\right)+\log (2)$$ is always positive. Moreover $f(1)=0$ and $f'(1)=1-\log(2)$.
Concerning the function $$g(q)=\frac 1 q - \Bigg[2^{-q}\sqrt{\pi } \frac{ \Gamma (q)}{\Gamma \left(q+\frac{1}{2}\right)} \Bigg]$$ you could easily show that its maximum is very close to $q=\sqrt {10}$ and $g(\sqrt {10})\sim \frac 15$
When $q > 2$, see @Gary's comment.
When $1 < q \le 2$, we have \begin{align*} \frac{\Gamma(q)\Gamma(q + 1)}{\Gamma(2q + 1)} &= \int_0^1 t^q (1 - t)^{q - 1}\mathrm{d} t \\ &= \int_0^{1/2} t^q (1 - t)^{q - 1}\mathrm{d} t + \int_{1/2}^1 t^q (1 - t)^{q - 1}\mathrm{d} t \\ &= \int_0^{1/2} t^q (1 - t)^{q - 1}\mathrm{d} t + \int_0^{1/2} (1 - t)^q t^{q - 1}\mathrm{d} t \\ &= \int_0^{1/2} t^{q- 1} (1 - t)^{q - 1}\mathrm{d} t\\ &\le \int_0^{1/2} t^{q- 1} (1 - (q - 1)t)\mathrm{d} t\\ &= \frac{-q^2 + 3q + 2}{2^{q + 1}q (q + 1)} \end{align*} where we have used Bernoulli inequality to get $(1 - t)^{q - 1} \le 1 - (q - 1)t$. Thus, we have $$q 2^q \frac{\Gamma(q)\Gamma(q + 1)}{\Gamma(2q + 1)} \le \frac{-q^2 + 3q + 2}{2 (q + 1)} < 1.$$
We are done.