Convexity or concavity implied by an inequality

Question

I have discovered the following fact:

Let $f(x)$ be a $C^3$ function on $[0;\infty[$ such that $$4|f'(x)|\geq f(x)^2.$$ With a Schwarzian derivative positive

If $f(x)$ is negative on $[0;\infty[$, then $f(x)$ is concave.
If $f(x)$ is positive on $[0;\infty[$, then $f(x)$ is convex.

I have no idea to prove this and I think it's hard.

Thanks a lot.

Answer 1

This is not true: Consider $$ f(x) = -\frac1{x+1}. $$ It is smooth, $f'= f^2$, thus $4f'\ge f^2$, $f'\ge0$, but it is not convex.

Answer 2

Consider$$ f(x) = \frac{1}{x^2 + x + 2}. \quad (x \geqslant 0) $$ On the one hand, for any $x > 0$,$$ 4|f'(x)| = 4 \left| -\frac{2x + 1}{(x^2 + x + 2)^2} \right| = \frac{4(2x + 1)}{(x^2 + x + 2)^2} \geqslant \frac{1}{(x^2 + x + 2)^2} = (f(x))^2. $$ On the other hand,$$ f''(x) = \frac{2x^2 + 2x - 3}{(x^2 + x + 2)^4}, $$ thus $f$ is not convex.