I want to prove that for all $n\in\mathbb{N}$ the following inequality holds
$$A_n(a)=\int_0^1t^{n-1/2}(1-t)^{-1/2}\left(\frac{1}{1-a}-t\right)^{-n}dt\hspace{0.15cm}>\pi\hspace{0.5cm},\hspace{0.5cm}\forall a\in[0,1/4]$$
The integral on the left side of inequality can be written as
$\displaystyle \,\,\,A_n(a)=\pi\sum_{k=0}^n\binom{n}{k}(-1)^{n-k}\int_0^1\frac{1}{\sqrt{t(1-t)}}\left(1-(1-a)t\right)^{-k}dt$
$\displaystyle \,\,\,A_n(a)=\pi\sum_{k=0}^n\binom{n}{k}(-1)^{n-k}{}_2F_1\left(1/2,k,1;1-a\right)$
$\displaystyle \,\,\,A_n(a)=\frac{(1-a)^n\sqrt{\pi }\,\, \Gamma\left(\frac{1}{2}+n\right)}{\Gamma(n+1)} {}_2F_1\left(n,\frac{1}{2}+n,1+n;1-a\right)$
In the 2 first cases is enough prove that sumatory, starting in $k=1$,is positive.
Other way: to prove that $A_n(a)$ is increasing (in $n$).
Any help will be welcome.