TITLE: Show that there is no differentiable function $f :\Bbb R \to\Bbb R$ such that $f(0) = 1$ and $f'(x) ≥ (f(x))^2\space\forall x\in\Bbb R$
QUESTION [7 upvotes]: Show that there is no differentiable function $f :\Bbb R \to\Bbb R$ such that $f(0) = 1$ and $f'(x) ≥ (f(x))^2\space\forall x\in\Bbb R$.

My attempt:
Suppose there is a differentiable function such that the above statement holds true.
Then, as the function is differentiable so it will satisfy the mean value theorem.
$$f(x)-f(0)=f'(c)(x-0),x>0,c \in (0,x)$$
$$f(x)-1=f'(c)x$$
$$f(x)-1 \ge (f(c))^2 x$$
$$f(x)-(f(c))^2 x -1 \ge 0$$
If we let $h(x) = f(x)-(f(c))^2.x-1$ then $h'(x) \ge 0$ then $f'(x)-(f(c))^2 \ge 0$ . How do I proceed after this and obtain a contradiction.

REPLY [7 votes]: Consider the function $g(x) = x+\frac{1}{f(x)}$.
Note $g'(x)= 1-\frac{f'(x)}{f^2(x)} \leq 0$. Hence $g(x)$ is a non-increasing function.
But $f'(x) \geq f^2(x) \geq 0$ implies $f(x)$ is non-decreasing function.
Since, $f(0) = 1$ implies $f(x) \geq 1 \geq 0$ (as $f$ is non-decreasing$).
Hence $g(100) \geq 100 $ but $g(x)$ is a non-increasing function implies $g(100) \leq f(0) = 1$. This a contradiction.