$\int^x_0\frac1{f'(t)}dt = \int^x_02f(t)dt \tag{1}$ 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 $\frac{d\left(\int^x_02f(t)dt\right )}{dx} =2f(x)$