Inspired by Sobolev inequalities, I would like to find a universal constant $C>0$ such that
$$ \left(\int_0^1 |f'(x)| dx\right)^2 \geq C \left(\int_0^1 f(x)dx -1\right) $$
for all functions $f \in C^1[0,1]$ satisfying $f>0$, $f(0)=f(1)$, $f'(0)=f'(1)$ and $\int_{0}^1\sqrt{f(x)}d x = 1$.
For the moment, when I assume $ |f_\max-f_\min| \geq 1$, then I have
$$ f(x) -1 \leq f(x_0)-1 + \int^1_0 |f'(x)|dx \leq \int^1_0|f'(x)|dx,$$
where $f(x_0) = \min_{x\in[0,1]}f(x) \leq 1$ and also $ \int^1_0|f'(x)|dx \geq 1$ thus
$$ \left(\int_0^1 |f'(x)| dx\right)^2 \geq \left(\int_0^1 |f'(x)| dx\right) \geq \int_0^1 f(x)dx -1.$$
However, I'm stuck at handling the case $|f_\max-f_\min| \leq 1$, I really need help here. My intuition is that, when is is "nearly" equal to 1, then both quantity are very small, but because of the condition $\int_0^1\sqrt{f(x)}dx = 1$, the quantity $\int_0^1|f'(x)|dx$ is still big...