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.



Proving two inequalities involving the gamma and digamma functions
By Fedor Petrov's comment, it is enough to show that 
\begin{equation}
 g(a,x):=\frac{\partial^2}{\partial^2 x}\,\psi(a/x)
\end{equation}
is increasing in $a>0$ for each $x>0$. We have 
\begin{equation}
 g(a,x)=f(u)/x^2, 
\end{equation}
where $u:=a/x$ and 
\begin{equation}
 f(u)=\psi''(u)u^2+2\psi'(u)u. 
\end{equation}
So, it is enough to show that $f'>0$ on $(0,\infty)$. 


By the [Gauss formula, Theorem 1.6.1](https://www.cambridge.org/core/books/special-functions/1F1C575CEA780EE774F5518C7963BF08)  

\begin{equation}
 \psi(u)=\int\_0^\infty\Big(\frac{e^{-z}}z-\frac{e^{-zu}}{1-e^{-z}}\Big)dz 
\end{equation}
for $u>0$, we have 
\begin{equation}
 f'(u)=\psi'''(u)u^2+4\psi''(u)u+2\psi'(u)
 =\int\_0^\infty U(z)(2 - 4 u z + u^2 z^2)e^{-zu} dz, \tag{1} 
\end{equation} 
where 
\begin{equation}
 U(z):=\frac{z}{1-e^{-z}}. 
\end{equation}
Integrating twice by parts, we have 
\begin{equation}
 f'(u)
 =\int\_0^\infty U'(z)z (-2 + u z)e^{-zu} dz
 =\int\_0^\infty U''(z)z^2 e^{-zu}dz>0, 
\end{equation}  

as desired. 
Here we used the fact that 
\begin{equation}
 U''(z)=\frac{e^{-2z} (2 + e^z (-2 + z) + z)}{(1 - e^{-z})^3}
 =\frac{e^{-2z} }{(1 - e^{-z})^3}\,\int\_0^z(z-t)e^t t\,dt>0
\end{equation}
for $z>0$. 


**Added in response to Fedor Petrov's comment on this answer:** The factor $2 - 4 u z + u^2 z^2$ in the integrand in (1) changes its sign (twice) when $z$ varies from $0$ to $\infty$. That does not allow us to obtain the desired (and, a priori, plausible) inequality $f'(u)>0$ right from (1). However, the hope was that, if we integrate the factor $(2 - 4 u z + u^2 z^2)e^{-zu}$ repeatedly in $z$ (in the sense of the indefinite integral), then the sign pattern of this factor will get sufficiently smoothed out and become a constant sign pattern. Fortunately, this worked, and we were also fortunate to have $U''>0$ on $(0,\infty)$. 


