where $f(x):[c,\infty]\to [0,A]$ is increasing, positive, $c>0$, and $f(c)=0$ (or, if it makes a difference, assume $f(c)>0$, but if we need this please state it). ($A$ is some positive constant. I.e. $f$ is bounded. The point of $c$ is to get $0$ out of the domain, to deal with what GReyes mentioned in a comment)
This is related to this question, but with the direction reversed.
Edit: and as a potential clarification, by "for all $x$" (in the title), I mean for all $x$ in the domain.
We still can't have $f(c)=0$ for the same reason as in the comment: there's no way for $f(c)>cf'(c)$, and by continuity it would continue to fail for values just to the right of $x=c$.
In any case, there's no reason that concavity should hold, and here's how to construct a counterexample: start with a function satisfying the hypotheses, say $f(x)=2-1/x$ on $[1,\infty)$. Then modify it by adding a small function with a small derivative but a very large second derivative near a point, say $$ g(x) = \begin{cases} 0, &\text{if } 1\le x\le 55, \\ 10(x-55)^2, &\text{if } 55\le x\le 55.01, \\ 10(55.02-x)^2, &\text{if } 55.01\le x\le 55.02, \\ 0, &\text{if } x\ge 55.02. \end{cases} $$ (Note that $g(x)\ge0$ and $|g'(x)|\le \frac15$, while $g''(x)$ can be as large as $20$.) Then $f(x)+g(x)$ still satisfies the hypotheses but is no longer concave.