Consider the following function for any $a, b > 0$
$$ \ g\left( a,b\right) = \frac{% 3\Gamma \left( 3b+1\right) }{\Gamma \left( \frac{1}{a}+3b+1\right) }-\frac{% 5\Gamma \left( 2b+1\right) }{\Gamma \left( \frac{1}{a}+2b+1\right) }+\frac{% 2\Gamma \left( b+1\right) }{\Gamma \left( \frac{1}{a}+b+1\right) } \ $$
I am interested in the region $(a,b)$ such that $g\left( a,b\right)\geqslant0$.
I know that this can be negative (for example for $a = 1$, $b < \frac{1}{3}$, this is negative). My conjecture is that for almost all $a, b$ couple the sign is positive. My main interest in the problem is when $a, b \geqslant 1$ and I conjecture that the inequality holds in this region. However, I do not have a general proof. Any help appreciated.