The title says it all. Let $y$ be a fixed positive real value. For any such $y$, is $\psi(x) - \psi(x + y)$ an increasing function of $x > 0$?
Here $\psi(x)$ is the digamma function, defined by:
$$\psi(x) = \Gamma'(x)/\Gamma(x)$$
where $\Gamma(x)$ is the Gamma function.
Let $f(x,y)=\psi(x)-\psi(x+y)$. Then, using the integral representation of $\psi(x)$ as given by
$$\psi(x)=\int_0^\infty \left(\frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}} \right) \,dt $$
we find that
$$f(x,y)=\int_0^\infty \left(\frac{e^{-(x+y)t}-e^{-xt}}{1-e^{-t}} \right) \,dt \tag 1$$
Differentiation under the integral sign we see that
$$\frac{\partial f(x,y)}{\partial x}=\int_0^\infty t\left(\frac{e^{-xt}-e^{-(x+y)t}}{1-e^{-t}} \right) \,dt \tag 2$$
For $y>0$, $\frac{\partial f(x,y)}{\partial x}>0$ and $f(x,y)$ is an increasing function of $x$.