Proving that a function $f:[0,\infty)\to[0,\infty)$ satisfying these conditions is necessarily non-decreasing I have a function $f: [0, \infty) \to [0, \infty)$ which is smooth. I also have that

*

*$f(0) = 0$

*$f'(0) > 0$

*$f''(x) \leq 0$, for all $x \in [0, \infty)$
It intuitively makes sense then that $f$ on $[0, \infty)$ must be strictly non-decreasing (since if it was ever decreasing, we would have to eventually "pull up" and contradict clause 3). I want to prove this, and I was able to derive contradictions if at some point the slope was negative. If at point $c$ in $(0, infty)$, $f'(c) < 0$, take some point $c + h$ further on in $[0, \infty)$, $h > 0$, and three cases occur:
a. $f(c + h) = f(c)$ -> got a contradiction
b. $f(c + h) > f(c)$ -> got a contradiction
But for $f(c + h) < f(c)$, this leads nowhere since it can still technically happen. I am at a complete loss on how to go forward. I was able to show that $f$ can never intersect $0$.
Edit: My main goal is to claim that $f(a) \leq f(b)$ for any $a < b$ on $[0, \infty)$

Take the function defined at $ [0,+\infty)$ by
$$(\forall x\ge 0)\;\; f(x)=x(1-x)$$
we have
$$f(0)=0$$
$$f'(0)=1>0$$
and
$$(\forall x\ge 0)\;\; f''(x)=-2\le 0$$
