Suppose that $\space f$ is positive and monotone increasing on $(0,\infty)$, $\space f \in AC[a,b]$(absolutely continuous) for every finite interval [a,b], and there is a constant $C>0$ such that $\space f(x)\leq Cx^2$ for all $x>0$. Prove that $\int_0^\infty1/f'=\infty$.
What I got is:
Fix $\space 0<a<b<\infty$, \begin{align} b-a=\int_a^b1=\int_a^bf'^{1/2}\cdot f'^{-1/2}&=|\langle f'^{1/2},f'^{-1/2}\rangle|\\ &\leq \|f'^{1/2}\|_2\cdot\|f'^{-1/2}\|_2 \end{align} Rearrange the inequality and take square of both sides to get: \begin{align} \|f'^{-1/2}\|_2^2 & \geq \frac{(b-a)^2}{\|f'^{1/2}\|_2^2}\\ \int_a^b f'^{-1} & \geq \frac{(b-a)^2}{\int_a^b f'}=\frac{(b-a)^2}{f(b)-f(a)} \end{align} And I got stuck here. I think the next step should be using the assumption $\space f(x)≤Cx^2$ to show that by letting $a\rightarrow0$ and $b\rightarrow \infty$, the integral will go infinity. But I don't know how.
Thanks!
Hint: By Jensen,
$$\frac{1}{x}\int_0^x \frac{1}{f'} \ge \frac{1}{\frac{1}{x}\int_0^x f'}.$$