Suppose $f(x)$ is a differentiable function on $\mathbb R$ with continuous derivative. For any $x \in \mathbb R$, $f’(x)>f(f(x))$. Prove that for any $x\ge 0$, $f(f(f(x)))\le 0$.
I don’t know how to use the continuity of the function’s derivative in this problem. The only thing I get right now is that $f(f(f(x)))\le f’(f(x))$ by substituting $f(x)$ into $x$, but I can’t prove $f’(f(x))\le 0$.
I know that it is an old question, but I came across it and I have an ansatz. First note that $$ (f\circ f)' = (f'\circ f)\cdot f' > (f\circ f\circ f)\cdot (f\circ f) $$ giving $$ f\circ f\circ f < \dfrac{(f\circ f)'}{f\circ f} = \big{(} \ln(\vert f\circ f\vert)\big{)}'. $$ That means that the question comes down to proving that $\vert f\circ f\vert$ is monotonely decreasing. In particular, $f\circ f$ needs to be either positive non-increasing or negative non-decreasing.
