# 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?

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ok, here is it.. the inequality is equivalent to $$\int_0^a \left(\frac{f(x)}{2x}\right)^2-f'(x)^2 \, dx \leq 0$$ First, searching for "equality" solution in the form $x^k$, we get $k=1/2$, but it fails. Also the "average" secant that is $$\frac{f(x)-f(0)}{x-0} = f'(\xi_x)$$ where $\xi_x \in (0,x)$. transform the integral to $$\int_0^a f'(c_x)^2-4f'(x)^2 \, dx$$ which "intuitively" saying that the integral of the difference of function $f'$ evaluated at $\xi_x \in (0,x)$ and four times evaluated at $x$ is negative.. I have a strong feeling this is true.. thanks. –  Ajat Adriansyah Jun 2 '11 at 4:37
I transform to this form, to see if there is a closely related inequality, for example "maybe" a "Chebishev-Like" Inequality $$\int_0^a g'(x) f'(x) \, dx \geq a^{-1} \int_0^a g'(x) \, dx \int_0^a f'(x)\, dx$$ with some additional requirement. –  Ajat Adriansyah Jun 2 '11 at 4:48
Did you try $f(x) = \sqrt{x} \log(x)$. –  Fabian Jun 2 '11 at 14:04
I think if you ask that $f'(x)$ is bounded then the inequality holds. You can check with the Euler-Lagrange equations that the integral assumes its minimum for $f(x) \propto \sqrt{x}$. –  Fabian Jun 2 '11 at 14:13
thanks Fabian, yes your example works.. and your suggestion too :) –  Ajat Adriansyah Jun 3 '11 at 3:00

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.

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great!! thanks! –  Ajat Adriansyah Jun 3 '11 at 16:57
I made the link to the Wikipedia page on Hardy's inequality clickable by replacing the ' in the link by its percent equivalent %27. I hope that's fine with you. –  t.b. Jun 3 '11 at 16:57
it's certainly fine with me.. I didn't know how to do it so I just left it as is. –  Zarrax Jun 3 '11 at 17:03
thanks Theo :) it's alright.. –  Ajat Adriansyah Jun 3 '11 at 17:03

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.

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