Let $f(x)=x \psi(x+1)$, where $\psi$ is the digamma function. Define 
$$g(x)=(f(ax)+f((1-a)x)-f(x))-(f(ax+by)+f((1-a)x+(1-b)y)-f(x+y)),$$
where $0\le a,b\le 1$ and $x,y\ge 0$.


How to show that $g(x)$ is increasing in $x$ on $[0,\infty)$? Thank you!




---


Here is what I have tried. Let $h(x,y)=f(ax+by)+f((1-a)x+(1-b)y)-f(x+y)$. Then we have $g(x)=h(x,0)-h(x,y)$. To show that $g(x)$ is increasing, it suffices to show that $$\frac{\partial}{\partial x}(h(x,0)-h(x,y))\ge 0. $$
Then it is sufficient to show that $$
\frac{\partial^2 h(x,y)}{\partial y \partial x}\le 0,
$$
which is equivalent to $$
f''(x+y)\ge abf''(ax+by)+(1-a)(1-b)f''((1-a)x+(1-b)y),
$$
where $f''(x)=2\psi\_1(1+x)+x\psi\_2(1+x)$, where $\psi\_1$ and $\psi\_2$ are the first and second derivatives of the digamma function $\psi$. Then if I use the series representation of polygamma functions
$$\psi\_m(z)=(-1)^{m+1}m!\sum\_{k\ge 0}\frac{1}{(z+k)^{m+1}},$$
the above inequality is equivalent to 
$$
\sum\_{k\ge 1}k\left[ \frac{1}{(x+y+k)^3}-\frac{ab}{(ax+by+k)^3}-\frac{(1-a)(1-b)}{((1-a)x+(1-b)y+k)^3} \right]\ge 0.
$$
I find that if $a,b,x,y$ are fixed, $\frac{1}{(x+y+k)^3}-\frac{ab}{(ax+by+k)^3}-\frac{(1-a)(1-b)}{((1-a)x+(1-b)y+k)^3} $ is negative for small $k$ and positive when $k$ is sufficiently large. So it seems that the inequality cannot be addressed term by term. I am thinking about if it can be proved by re-arranging the terms in the series and grouping them into positive groups. I also tried the similar strategy by using the integral representation of polygamma functions $$
\psi\_m(z)=(-1)^{m+1}\int\_0^\infty \frac{t^m e^{-zt}}{1-e^{-t}}dt,
$$
but it did not work either. 



A challenging inequality that involves the digamma function and polygamma functions
Max reduced the problem to proving that $(xf'')'\geqslant 0$. We have $f=x\psi(x+1)=1+x\psi(x)$, so $f''=(x\psi(x))''=x\psi''+2\psi'$, $f'''=x\psi'''+3\psi''$,$(xf'')'=xf'''+f''=x^2\psi'''+4x\psi''+2\psi'$. Exactly this guy is proved to be non-negative in the recent clever [MO answer](https://mathoverflow.net/questions/322293/proving-two-inequalities-involving-the-gamma-and-digamma-functions/322331#322331) by Iosif Pinelis to a different question. I wonder whether this is coincidence.


