Inequality for expected value of X and $X^{-1}$

Question

I've got the following problem: we are given a random variable X such that $0 < a \leq X \leq b$ and we need to estimate $(\mathbb{E}X)\mathbb{E}(X^{-1})$ from above. I suggest that $$(\mathbb{E}X)\mathbb{E}(X^{-1}) \leq \frac12\left(\frac{a}{b} + \frac{b}{a}\right)$$ I just guessed it but I'm quite sure that's true because I can prove it if $X$ can be only $c_1$ and $c_2$ with probabilities $p$ and $(1 - p)$ correspondingly. I'll be glad for any help.

Answer 1

If $\mu$ is the measure correponding to the density of $X$, then we can write both \begin{align*} (\Bbb EX)(\Bbb EX^{-1}) &= \int_a^b t\,d\mu(t) \int_a^b \frac1u\,d\mu(u) \\ (\Bbb EX)(\Bbb EX^{-1}) &= \int_a^b u\,d\mu(u) \int_a^b \frac1t\,d\mu(t); \end{align*} adding these together yields $$ 2 (\Bbb EX)(\Bbb EX^{-1}) = \int_a^b \int_a^b \bigg( \frac tu+\frac ut \bigg) \,d\mu(u)\,d\mu(t). $$ The fact that $g(y)=y+\frac1y$ is an increasing function for $y\ge1$ implies in particular that $\frac tu+\frac ut \le \frac ab+\frac ba$ when $a\le t\le b$ and $a\le u\le b$; therefore $$ 2 (\Bbb EX)(\Bbb EX^{-1}) \le \bigg( \frac ab+\frac ba \bigg) \int_a^b \int_a^b \,d\mu(u)\,d\mu(t) = \frac ab+\frac ba. $$