If I know that for $f$ the following always holds: $$f(\pmb{x}) > 0, \pmb{x} \in [0,1]^d$$ $$\int_{[0,1]^d}f(\pmb{x})\,d\pmb{x} = 1$$
I am interested for what class of functions the following also holds: $$\int_{[0,1]^d}\frac{1}{f(\pmb{x})}\,d\pmb{x} = 1$$
Clearly if $f$ is constant ($f = 1$) then the first statement holds, since:
$$f(\pmb{x}) = 1 \Rightarrow \frac{1}{f(\pmb{x})} = f(\pmb{x})$$
Are there more functions for which:
$$ \int_{[0,1]^d}\frac{1}{f(\pmb{x})}\,d\pmb{x} = \int_{[0,1]^d}f(\pmb{x})\,d\pmb{x}$$
$\phi (x)=\frac 1 x$ is a strictly convex function on $(0,\infty)$. $\phi (\int f) \leq \int \phi(f)$ which means $\frac 1 {\int f} \leq \int \frac 1 f$ and equality holds only when $f$ is a constant. The constant has to be $1$ of course.