Let $\Gamma$ denote the gamma function and suppose $a, b > 1$. Is the function defined as,
$f_b(a) = \cfrac{\Gamma(ab)}{\Gamma(ab + 1/2)} \cfrac{\Gamma(a + 1/2)}{\Gamma(a)}$,
increasing in $a$? I was asked to prove this formally, but have run into some difficulty. Most of my attempts so far have been some variation of using the digamma function, but I haven't been able to get anywhere meaningful.
We have that $\log\Gamma$ is a convex function by the Bohr-Mollerup theorem, and $f_b(a)$ is increasing with respect to $a$ iff $$\log f_b(a) = \log\frac{\Gamma(ab)}{\Gamma(a)}-\log\frac{\Gamma(ab+1/2)}{\Gamma(a+1/2)}=\int_{a}^{ab}\psi(x)-\psi(x+1/2)\,dx $$ is increasing with respect to $a$, where $$\psi(x)-\psi(x+1/2)=\sum_{n\geq 0}\left(\frac{1}{n+x+\frac{1}{2}}-\frac{1}{n+x}\right) $$ leads to: $$ \log f_b(a) = -\frac{1}{2}\int_{a}^{ab}\sum_{n\geq 0}\frac{dx}{(n+x)(n+x+1/2)} $$ and through such a representation it is straightforward to prove that $f_b(a)$ is an increasing function.