I am asking for your help with an exercise from my general physics I class. The exercise is about the Gamma function and reads in this way.
Let $b\in\mathbb{N}, b\ge 2$ and $a\in (0, 1)$. Prove that there exists an lower bound which does not depend on $a$ for $$\frac{\Gamma((a+b)/2)}{\Gamma((b-a)/2)}\cdot \left(\frac{\Gamma(b/2)}{\Gamma(b)}\right)^{a/b},$$ i.e. that there exists a positive constant $c$, non depending on $a$, such that $$\frac{\Gamma((a+b)/2)}{\Gamma((b-a)/2)}\cdot \left(\frac{\Gamma(b/2)}{\Gamma(b)}\right)^{a/b}\ge c.$$
Here $\Gamma$ is defined, for $Re(z)>0$, as $$\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} dt$$ (Note that here $a, b, a-b, a+b$ are all real and nonnegative numbers).
The only idea I have so far is to split $\int_0^\infty \dots = \int_0^1 \dots +\int_1^\infty \dots$. This led me to a huge number of integrals which I do not know how to manage and, honestly, I do not think is the right way to prove the result. Anyone could please help in proving this result?
Finally, just out of my curiosity, I would like to understand if one can get an upper bound, too (but this second thing is not important and this thing will not affect the acceptance of the answer).
Lower bound. By Gautschi’s Inequality $(5.6.4)$, \begin{align*} \frac{{\Gamma \big( {\frac{b}{2} + \frac{a}{2}} \big)}}{{\Gamma \big( {\frac{b}{2} - \frac{a}{2}} \big)}} & = \frac{{b - a}}{2}\frac{{\Gamma \big( {\frac{{b - a}}{2} + a} \big)}}{{\Gamma \big( {\frac{{b - a}}{2} + 1} \big)}} \ge \frac{{b - a}}{2}\left( {\frac{{b - a}}{2} + 1} \right)^{a - 1} \\ & \ge \frac{b}{2}\left( {\frac{b}{2} + 1} \right)^{a - 1} \ge \frac{1}{2}\left( {\frac{b}{2} + 1} \right)^a \end{align*} for any $b\ge 2$ and $0<a<1$. We also have $$ \left( {\frac{{\Gamma (b/2)}}{{\Gamma (b)}}} \right)^{1/b} = \left( {\sqrt \pi \,2^{1 - b} \frac{1}{{\Gamma ((b + 1)/2)}}} \right)^{1/b} \ge \frac{1}{2}\left( {\frac{1}{{\Gamma (b)}}} \right)^{1/b} \ge \frac{1}{2}\frac{1}{b} $$ if $b\ge 2$. Thus, \begin{align*} \frac{{\Gamma \big( {\frac{b}{2} + \frac{a}{2}} \big)}}{{\Gamma \big( {\frac{b}{2} - \frac{a}{2}} \big)}}\left( {\frac{{\Gamma (b/2)}}{{\Gamma (b)}}} \right)^{a/b} & \ge \frac{1}{2}\left( {\left( {\frac{b}{2} + 1} \right)\left( {\frac{{\Gamma (b/2)}}{{\Gamma (b)}}} \right)^{1/b} } \right)^a \\ & \ge \frac{1}{2}\left( {\left( {\frac{b}{2} + 1} \right)\frac{1}{2}\frac{1}{b}} \right)^a \ge \frac{1}{2}\left( {\frac{1}{4}} \right)^a \ge \frac{1}{8}. \end{align*} Upper bound. The quantity is unbounded. Assume that $0<a<1$. By the asymptotics $(5.11.12)$, \begin{align*} \frac{{\Gamma \big( {\frac{b}{2} + \frac{a}{2}} \big)}}{{\Gamma \big( {\frac{b}{2} - \frac{a}{2}} \big)}} \sim \left( {\frac{b}{2}} \right)^a \end{align*} as $b\to+\infty$. By Stirling's formula $$ \left( {\frac{{\Gamma (b/2)}}{{\Gamma (b)}}} \right)^{1/b} \sim \sqrt {\frac{{\rm e}}{2}} \frac{1}{{b^{1/2} }} $$ as $b\to+\infty$. Consequently, $$ \frac{{\Gamma \big( {\frac{b}{2} + \frac{a}{2}} \big)}}{{\Gamma \big( {\frac{b}{2} - \frac{a}{2}} \big)}}\left( {\frac{{\Gamma (b/2)}}{{\Gamma (b)}}} \right)^{a/b} \sim \left( {\frac{{\rm e}}{8}} \right)^{a/2} b^{a/2} $$ as $b\to+\infty$.