I wish to prove the following inequality: $$A \leq B$$ where $$A=\Gamma(1+\frac{2}{\beta})\cdot \frac{q^{\frac{1}{\beta}}}{1+\beta}\cdot {}_2F_1(1,1+\frac{2}{\beta};2+\frac{1}{\beta};1-q) $$ $$B=(\Gamma(1+\frac{1}{\beta}))^2\cdot \frac{(1-q^{\frac{1}{\beta}})}{1-q}$$ the parameters $q$ and $\beta$ satisfy following conditions:
$\beta \geq 1$,
$0 \leq q \leq 1$,
$\Gamma(x)$ the gamma function
and ${}_2F_1(a,b;c;z) = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}$ the hypergeometric function
The maximum of the quotient $\frac{A}{B}$ is obtained when $\beta=1$, i then tried unsuccessfully to express it by the product of 2 functions bounded by 1. Also it seems difficult to have the derivative as $\beta$ is in the F. Perhaps the hypergeom can simplify with the gamma function?
Any help would be appreciated.