Integral inequality

Question

Let $f$ be a continuously differentiable real-valued function on $[0,b]$, where $b>0$, with $f(0)=0$. Prove that $$\int\limits_0^b\frac{f(x)^2}{x^2}dx\leq4\int\limits_0^b f'(x)^2dx.$$ Thank you!

Answer 1

First, It follows from the differentiability that $\mathrm{LHS}$ converges (why?).

Now integrate by part (how?) and apply Cauchy-Schwarz inequality, we have

$$\int_0^b\frac{f(x)^2}{x^2}dx=-\frac{f(b)^2}b+2\int_0^b\frac{f(x)f'(x)}xdx\le 2\,\left(\int_0^b\frac{f(x)^2}{x^2}dx\int_0^bf'(x)^2dx\right)^{1/2}$$

The required inequality follows.