I'm having trouble proving the following inequality:

$$\forall p>1 \quad \forall m\geq 0 \quad \dfrac{m^2\Gamma(\dfrac{2m}{p})\Gamma(\dfrac{2m}{q})}{\Gamma(\dfrac{2m+2}{p})\Gamma(\dfrac{2m+2}{q})}\geq\dfrac{1}{4}p^2(p-1)^{\frac{2}{p}-2},$$ where as usual $q=\dfrac{p}{p-1}$. In fact, it seems clear from Mathematica that for a fixed $p$, the LHS is a decreasing function of $m$ (strictly unless $p=2$, in which case it's constant). The RHS can be seen to be the limit as $m\to \infty$. I actually only care about integer $m\geq 0$, but I don't find that helpful.

I have tried both a direct approach (three known inequalities that are nice enough to apply here, but lead to wrong inequalities) and working with the derivative, which naturally involves instances of the digamma function. Proving that the LHS is decreasing is equivalent to the following inequality: $$\forall p>1 \quad \forall m\geq 0 \quad \dfrac{1}{m}+\dfrac{1}{p}(\psi(\dfrac{2m}{p})-\psi(\dfrac{2m+2}{p}))+\dfrac{1}{q}(\psi(\dfrac{2m}{q})-\psi(\dfrac{2m+2}{q}))\leq0,$$ which again seems to be correct (if you're wondering, the limit as $m\to 0$ is negative for $p\neq2$). Much like before, I tried using two inequalities (for the digamma function), as well as the series representation. They seemed promising at first, but the inequalities gave me positive upper bounds, while the series converges too slowly to be useful (I suspect that any partial sum is positive for large enough $m$).

Any advice would be much appreciated. I'll be glad to explain more about the inequalities I've tried if requested.