Asking for general form of Integral Inequality of this kind Let $f\in C^1[0,a]$ and $f(0)=0$. Is it true that
$$\int_0^a \left(\sqrt{x}f(x)\right)^{\prime} \left(\frac{f(x)}{\sqrt{x}}\right)^{\prime}\, dx\geq 0\;\;?$$
What is the general form of this type inequality?
 A: I have found a particular proof for this (which is stronger since i proved $\int_0^a f'(x)^2\, dx \geq \int_0^a \left(\frac{f(x)}{2x}\right)^2 \, dx$).
since $f(0)=0$ we have $\lim_{x\rightarrow 0} \frac{f(x)^2}{x}=\lim_{x\rightarrow 0} 2f'(x)f(x)=2f'(0)f(0)=0$.
so integration by part gives us $\int_0^a \left(\frac{f(x)}{2x}\right)^2 \, dx + \frac{f(a)^2}{4a}= \int_0^a \frac{f'(x)f(x)}{2x}$.
By AM-GM we have
$$f'(x)^2 + \frac{f(x)^2}{4x^2} \geq \frac{f(x)f'(x)}{x}$$
integrate this inequality from $[0,a]$ and using $2 \int_0^a \left(\frac{f(x)}{2x}\right)^2 \, dx + \frac{f(a)^2}{2a} = \int_0^a \frac{f'(x)f(x)}{x}\, dx$ we are done.
A: In the form $\int_0^a \big(({f(x) \over 2x})^2 - (f'(x))^2\big) \, dx \leq 0$ this follows readily from Hardy's Inequality for $p = 2$. See http://en.wikipedia.org/wiki/Hardy%27s_inequality for this and generalizations.
To apply it, let your $|f'(x)|$ be what is called $f(x)$ in that article. Also you need to extend $f'(x)$ to be zero for $x \geq a$ to apply the theorem.
