I am trying to looking into function $f$ which satisfies the inequality $f(E[X]) - E[f(X)] > 0$ where $E[X]$ is the expectation of positive random variable $X$.
My questions is, what is the necessary and sufficient condition of $f$ satisfying the inequality?
Guess and Attempt:
Do not know how to attack the problem rigorously.
I have tried a case that $f(x) = \ln(x)$, and I found that the inequality $f(E[X]) - E[f(X)] > 0$ is exactly A.M. > G.M..
For $f(x) = x^n$ with $n > 1$, $f(E[X]) - E[f(X)] = (E[X])^n - (M_n[X])^n < 0$ because it is the difference between generalized mean $M_n$ and mean $E$, both to the power $n$.
For $f(x) = x$ the equality holds obviously.
So I guess that $f(x) < x$ for positive $x$ is the required necessary and sufficient condition.