Approximating $\mathbb{E}[1/X]$ I am well aware (as for instance discussed here https://math.stackexchange.com/questions/910846/is-it-true-in-general-that-e1-x-1-ex) that for an arbitrary random variable $X$ it does not hold that $\mathbb{E}[1/X]=1/\mathbb{E}[X]$. However, when one cannot calculate  $\mathbb{E}[1/X]$ directly, is there a good (rigorous) way to approximate $\mathbb{E}[1/X]$ (ideally, regardless of the distribution of $X$)?
 A: Assume that $X\ge0$. Then, by Jensen's inequality for the convex function $x\mapsto1/x$, 
\begin{equation}
 E\frac1X\ge\frac1\mu,
\end{equation}
where $\mu:=EX$. Thus, we have the lower bound $\frac1\mu$ on $E\frac1X$; this bound is exact, as it is attained when $X$ is a constant. 
On the other hand, it is easy to see that there is no finite upper bound on $E\frac1X$ in terms of $\mu$ only. Indeed, for any real $\mu>0$, let $P(X=0)=1/2=P(X=2\mu)$. Then $E\frac1X=\infty$ while $EX=\mu$. Even if we assume that $X>0$, still there is no finite upper bound on $E\frac1X$ in terms of $\mu$ only. Indeed, for any real $\mu>0$ and any $a\in(0,\mu)$, let $P(X=a)=1/2=P(X=2\mu-a)$. Then $EX=\mu$ while $E\frac1X=\frac1{2a}+\frac1{2(2\mu-a)}\to\infty$ as $a\downarrow0$.
If $X$ is known to be bounded away from $0$, so that $X\ge a$ for some real $a>0$, then trivially $1/X\le1/a$ and hence 
\begin{equation}
E\frac1X\le \frac1a.  
\end{equation}
The latter upper bound on $E\frac1X$ is exact here for each feasible value of $\mu=EX$, even though this bound does not even involve $\mu$. Indeed, take any real $\mu\ge a$. For any real $u>\mu$, let $P(X=u)=p=1-P(X=a)$, where $p:=\frac{\mu-a}{u-a}\in[0,1)$, so that $EX=\mu$. On the other hand, letting $u\to\infty$, we have $p\to0$ and hence $E\frac1X=\frac1a\,(1-p)+\frac1u\,p\to\frac1a$, so that the upper bound $\frac1a$ on $E\frac1X$ is attained in the limit. 
Finally, if $X$ is known to be bounded away both from $0$ and $\infty$, so that $a\le X\le A$ for some positive real $a,A$ such that $a<A$, then, by the convexity of the function $x\mapsto1/x$, we have 
$\frac1X\le\frac{A+a-X}{Aa}$ and hence 
\begin{equation}
 E\frac1X\le\frac{A+a-\mu}{Aa}. 
\end{equation}
The latter bound on $E\frac1X$ is again exact, as it is attained when $P(X=A)=p=1-P(X=a)$, where $p:=\frac{\mu-a}{A-a}\in[0,1]$, so that $EX=\mu$.
