Is such an integral inequality possible? Let $X$ be measure space of finite measure. Let $f:X \to \mathbb{R}$ be positive,bounded and measurable, and $p \le 1$.
I would like to have the following inequality:
$$
\int_X f^p \ge F \big(\int_X f\big)
$$
where $F:\mathbb{R}^+ \to \mathbb{R}^+$ can be any function, which does not depend on the pointwise bound on $f$. (It can depend on properties of $X$, such as its measure).
Holder's inequality works in the wrong direction... 
Is there any chance for such a thing?
 A: For $p = 1$, $F(t) = t$ works. Hence we assume $0 < p < 1$ in the following.
If $X$ has subsets of arbitrarily small positive measure, then we can find a sequence with $\int_X f_n = 1$ for all $n$, but $\int_X f_n^p \to 0$ and hence such an $F$ cannot exist. For if $\mu(A_n) > \mu(A_{n+1}) \to 0$, let
$$f_n = \frac{n-1}{n\mu(A_n)} \chi_{A_n} + \frac{1}{n\mu(X\setminus A_n)} \chi_{X\setminus A_n}.$$
Then $\int_X f_n = 1$, and
$$\int_X f_n^p = \biggl(\frac{n-1}{n}\biggr)^p \cdot \mu(A_n)^{1-p} + \frac{\mu(X\setminus A_n)^{1-p}}{n^p} \leqslant \mu(A_n)^{1-p} + \frac{\mu(X)^{1-p}}{n^p} \to 0.$$
So we need a purely atomic measure space whose atoms have measure bounded below by a strictly positive constant. That means there are only finitely many atoms if $X$ shall have finite measure. And in that case, we can find such an $F$. Let $c$ be the smallest measure of an atom. Then $F(t) = \min \: \{t,c\}$ satisfies the inequality. If $f \leqslant 1$, then we have
$$\int_X f^p \geqslant \int_X f \geqslant F\biggl(\int_X f\biggr).$$
And if $f(x) \geqslant 1$ on some atom, then
$$\int_X f^p \geqslant c \geqslant F\biggl(\int_X f\biggr).$$
