TITLE: $\int \frac{1}{f’(x)}$ diverges
QUESTION [8 upvotes]: I got this question in a prelims exam. 
$f$ is a positive, continuously differentiable function on $(0,\infty)$ such that $f’ >0$. If $f(x) \leq Cx^2$ for some $C>0$ then $\int \frac{1}{f’(x)}$ diverges.
There is a hint in the problem which asks to establish that if $\int \frac{1}{f’(x)}< \infty$ then $\int\limits_{a}^{\infty} \frac{1}{f’(x)} \to 0$ as $a\to \infty$. I can establish the claim made in the hint by writing $\int f = \sum_{n} \int_{n}^{n+1} f$ and then observing that the above sum converges absolutely and therefore the tail of the series (which is the required integral) goes to 0. 
I believe that I am now supposed to show that in our case $\int_{a}^{\infty} \frac{1}{f’(x)}$ does not go to zero as $a\to \infty$. But, I am unable to make use of the fact that $f(x)\leq Cx^2$. I do not know if it is useful to note that $\frac{1}{f’(x)}=(f^{-1})’(f(x))$.
Any help would be highly appreciated. I would like some hint on how to use this condition.

REPLY [7 votes]: Note that by Cauchy-Schwarz inequality it holds
$$
\int_A^{2A} \frac {dx} {f'(x)}\cdot \int_A^{2A} f'(x)dx \ge \left(\int_A^{2A} 1 \ dx\right)^2 = A^2.
$$ We can also see that
$$
4CA^2 \ge f(2A) > f(2A) - f(A) = \int_A^{2A} f'(x) dx
$$ by the fundamental theorem of calculus. This gives
$$
\int_A^{2A} \frac {dx}{f'(x)} \ge \frac 1{4C},\quad \forall A>0,
$$ which implies
\begin{align*}
\int_0^\infty \frac {dx}{f'(x)} =& \sum_{j=-\infty}^\infty \int_{2^j}^{2^{j+1}} \frac {dx}{f'(x)} \\
\ge & \sum_{j=-\infty}^\infty \frac 1{4C} = \infty.
\end{align*}