TITLE: Proving $\int_0^1\frac{\vert f(x)\rvert^2}{x^2}\,\mathrm dx\le4\int_0^1{\vert f'(x)\rvert^2}\,\mathrm dx$ when $f\in\mathcal C^1([0,1])$ and $f(0)=0.$
QUESTION [14 upvotes]: Let us assume $f \in \mathcal{C}^1([0,1])$ and $f(0)=0.$ Prove that $$\int_{0}^{1} \frac{\vert f(x) \rvert^2}{x^2}dx \le 4 \int_{0}^{1} {\vert f'(x) \rvert^2}dx.$$
By integrating by parts I obtained the following 
$$\int_{0}^{1} \frac{\vert f(x) \rvert^2}{x^2}dx = -\frac{1}{x} \lvert f(x) \rvert^2\Big|_0^1+2\int_{0}^{1} \frac{f(x)|f'(x)|}{x |f(x)|} \le 2\int_{0}^{1} \frac{|f'(x)|}{x } $$
but I'm not sure of the result and its usefulness, it's easy Calculus, but I can't go on. Any suggestions?

REPLY [6 votes]: You have almost done it with your integration by parts attempt, just missing one simple last step. Again as you started was
\begin{align}
I&=\int_0^1\frac{f(x)^2}{x^2}\,dx=\underbrace{-\frac{f(x)^2}{x}\Bigg|_0^1}_{\le 0}+2\int_0^1\frac{f(x)}{x}f'(x)\,dx\le
2\int_0^1\frac{f(x)}{x}f'(x)\,dx\le\\
&\le 2\left(\int_0^1\frac{f(x)^2}{x^2}\,dx\right)^{1/2}\left(\int_0^1 f'(x)^2\,dx\right)^{1/2}=2\sqrt{I}\left(\int_0^1 f'(x)^2\,dx\right)^{1/2}.
\end{align}
Now divide by $\sqrt{I}$ and square all.