Expectation on 1/X

It is very rarely true. Let's do a random example. Let $X$ be uniform on $[1,3]$. Then $E(X)=2$. But $$E\left(\frac{1}{X}\right)=\int_1^3 \frac{1}{x}\cdot\frac{1}{2}\,dx=\frac{\log 3}{2}\ne \frac{1}{2}.$$

For a simpler example, let $X=1$ with probability $1/2$, and let $X=3$ with probability $1/2$. Then $E(X)=2$.

But $E(1/X)=(1/2)(1)+(1/2)(1/3)=2/3$.

Jensen's inequality for functions of RVs is $\mathbf{E} \varphi(x) \geq \varphi(\mathbf{E}X)$ when $\varphi$ is convex and $\mathbf{E} \varphi(x) \leq \varphi(\mathbf{E}X)$ for concave functions. Clearly $Y = \frac{1}{X}$ is a convex function when restricted to the positive reals or the negative reals, so ${\bf E}[X^{-1}]\ge {\bf E}[X]^{-1}$ is true as long as $X$ is almost surely positive or almost surely negative.

For such a case, it is a good idea to study Jensen's inequality.

Another counterexample to the one given by André Nicolas is this one. Consider $X$ to be a normal distribution with mean $\mu$ and variance one. Then $E[X]=\mu$ but not only is $E[\frac{1}{X}]$ not in general equal to $1/\mu$; rather, it does not exist.