Is this monotonicity property equivalent to convexity?

$\psi$ can be non-concave. Here is a counterexample.

Let $$\psi(r) = \frac{r(r^2+2)}{r^2+1}.$$ Clearly, $\psi$ is smooth, and $\psi(0)=0$. Also, $\psi$ is strictly increasing since $$\psi'(r) = \frac{r^4+r^2+2}{(r^2+1)^2} > 0.$$

We have $$f(r) = \psi'(r) + \frac{\psi(r)}{r} = \frac{2(r^4+2r^2+2)}{(r^2+1)^2}$$ and $$f'(r) = - \frac{8r}{(r^2+1)^3}.$$ Thus, $f$ is non-increasing.

However $\psi$ is not concave since $$\psi''(r) = \frac{2r(r^2-3)}{(r^2+1)^3}.$$ For example, $\psi''(2) > 0$.