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.