Prove there exists no $f(x)$ with $ f'(x)\geq f^{\alpha}(x)$.

Question

Let $f(x)$ be differentiable over $[0,+\infty)$ and $f(x)> 0$. Given $\alpha>1$, prove that there exists no such $f(x)$ that $ f'(x)\geq f^{\alpha}(x)$ over $[0,+\infty).$

Obviously, it suffices to show that there exists a $c \in [0,+\infty)$ such that $f'(c)<f^{\alpha}(c).$ For this purpose, denote $F(x)=\dfrac{f^{1-\alpha}(x)}{1-\alpha}$, then we only need to prove $F'(c)<1$ for some $c\in [0,+\infty)$. This will work?

Answer 1

Suppose that $f>0$ satisfies $f'(x)\geq f^\alpha(x)$ over $[0,+\infty)$ with $\alpha>1$. Let $$F(x)=x-\frac{f^{1-\alpha}(x)}{1-\alpha},\ \ x>0.$$ Then $F'(x)=1-\frac{f'(x)}{f^\alpha(x)}\leq 0$ and thus $F$ is non-increasing over $[0,+\infty)$. Note that $\alpha>1$ implies that $F(x)>x$ so $\lim_{x\to+\infty}F(x)=+\infty$, a contradiction.

Answer 2

I guess we call this a hint. For a fixed positive integer $n,$ define $$ g(x) = \left( \frac{n}{n-x} \right)^n $$ Calculate $g'.$ What is $g(0) \; ? \; $ What ODE does $g$ solve? Is $g$ defined for all nonnegative $x\; ? $