# Functional Inequality question where $\int^x_0\frac1{f'(t)}dt = \int^x_02f(t)dt$ ,$0 \leq x \leq 1$ and $f(0) = 0$

$$\int^x_0\frac1{f'(t)}dt = \int^x_02f(t)dt \tag{1}$$ where $0 \leq x \leq 1$ and $f(0) = 0$

I need to prove that $$f(\frac1{\sqrt{2}})> \frac1{\sqrt{2}}$$

$$f(\tan (x))> \tan(x) > x , x \in (0,\frac{\pi}{4})$$ $$f(e^{-x^2})\geq e^{-x^2}$$

The problem is that i dont know how to derive the function it self.

All I could do was say that $$\frac{d\left(\int^x_0\frac1{f~'(t)}dt)\right )}{dx} = \frac1{f~'(x)}$$

and $$\frac{d\left(\int^x_02f(t)dt\right )}{dx} =2f(x)$$

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I was wondering: assume that $\int_0^x g(t)\, dt=0$ for every $x \geq 0$. Can we deduce anything about $g$? – Siminore Jul 5 '12 at 14:13
There is no functional inequality here. – Did Jul 5 '12 at 15:07

You get

$$\frac{1}{f'(x)}=2f(x)$$

Thus

$$2f'(x)f(x)=1 \,.$$ or

$$\left( f(x)^2 \right)' =1 \,.$$

Integrate, find the constant and you are done.

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but differentiating two equal integrals doesn't necessarily make them equal right ? – The-Ever-Kid Jul 5 '12 at 14:28
If two functions are equal, their derivatives are also equal ;) – N. S. Jul 5 '12 at 14:28
Always ?....... – The-Ever-Kid Jul 5 '12 at 14:29
Well if two functions are equal, they are the SAME function... – N. S. Jul 5 '12 at 14:29
How is the derivative of a function defined, @The-Ever-Kid? It is defined in turns of the values of the function. If two functions are equal at all points, then of course their derivatives are equal. – Thomas Andrews Jul 5 '12 at 14:30