Let $X$ be a random variable. Is it ever true that $$E\left({\frac 1 X}\right) \stackrel{?}{=} \frac 1 {E(X)} \text{ ?}$$
I'll assume $X$ never takes on the value of $0$.
I'll use the notation for discrete RV's, otherwise replace $\sum$ with $\int$. Let $f$ be the mass/density function.
$$E\left({\frac 1 X}\right) = \sum \frac 1 x f(x)$$
$$\frac 1 {E(X)} = \frac 1 {\sum x f(x)}$$
Assume they're equal. Then:
$$\sum \frac 1 x f(x) = \frac 1 {\sum x f(x)}$$
$$\implies \left(\sum \frac 1 x f(x)\right)\left(\sum xf(x)\right) = 1$$
This is a type of convolution, which would mean that $$\dfrac {f(x)}{x} = (x f(x))^{-1}$$
in regards to this convolution. Is that possible? Do functions have inverses under convolution?
By Jensen's inequality, for a positive-valued random variable $X$ we have
$$E\left(\frac{1}{X}\right)\ge \frac1{E(X)}$$
If $X$ is negative-valued, then the inequality is reversed.
This is because the map $x\mapsto \frac{1}{x}$ is convex for $x>0$ and concave for $x<0$. Equality holds precisely when $X$ is constant with probability $1$.
However equality can hold if we do not restrict ourselves to either positive-valued or negative-valued random variables. As a non-trivial example, consider a discrete random variable $X$ with $$P(X=-1)=\frac{1}{9}\,,\, P\left(X=\frac{1}{2}\right)=P(X=2)=\frac{4}{9}$$
Then indeed $$E\left(\frac{1}{X}\right)=\frac{1}{E(X)}=1$$
For details, see https://stats.stackexchange.com/q/305713/119261.