TITLE: Is this monotonicity property equivalent to convexity?
QUESTION [7 upvotes]: This is a follow-up of this question.
Let $\psi:[0,\infty) \to [0,\infty)$ be a strictly increasing  $C^2$ (or $C^{\infty}$) function, satisfying $\psi(0)=0$. 
Suppose that the function 
$f(r)=\psi'(r)+\frac{\psi(r)}{r}$ is non-increasing. 

Must $\psi$ be concave?

The converse statement is true, i.e. $\psi$ concave implies $f$ non-increasing: Indeed,
$$
f'(r)=\psi''(r)+\frac{1}{r}(\psi'(r)-\frac{\psi(r)}{r}),
$$ $\psi'' \le 0$ by concavity, and since $\psi(r)=\int_0^r \psi'(t)dt \ge \int_0^r \psi'(r)dt=r\psi'(r)$, the term $\frac{1}{r}(\psi'(r)-\frac{\psi(r)}{r})$ is also non-positive.
Edit:
Here is a partial result-I can prove that $\psi''(0) \le 0$.
By our assumption
$$
0 \ge f'(r)=\psi''(r)+\frac{1}{r}(\psi'(r)-\frac{\psi(r)}{r}),
$$
for every $r>0$. Using the mean value theorem (twice), we can rewrite this as $$
f'(r)=\psi''(r)+\psi''(s(r)) \le 0, \tag{1}
$$
where $s(r)$ is some point in $(0,r)$.
In particular, taking the limit when $r \to 0$, we deduce that $\psi''(0) \le 0$.

REPLY [2 votes]: $\psi$ 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, $\psi''(2) > 0$.