The best possibility of the bound for the Kantorovich inequality and some remarks (Q1372378)

Let \((\Omega,{\mathcal F},p)\) be a probability measure space. For any complex-valued random variable \(X\) with \(0< m\leq|X|\leq M\), a.s., the following inequality holds \[ \Biggl|\int X dp \int {1\over X} dp\Biggr|\leq {(m+ M)^2\over 4mM}. \] The equality is attained iff there exists \(0\leq\theta\leq 2\pi\) such that \[ p(\{X= me^{i\theta}\})= p(\{X= Me^{i\theta}\})= \begin{cases} {1\over 2}, &\text{if }m<M,\\ 1, & \text{if }m= M.\end{cases} \] {}.