The function $1/x$ is convex on the interval $x > 0$ and two times differentiable. So, for $x > 0$, Jensen's inequality implies
$$E(1/x) \geq 1/E(x).$$
But for $x < 0$, $1/x$ is not convex, but concave. However, $1/(-x)$ is convex and two times differentiable. Let $g(x) = f(-x) = -1/x$. Then
\begin{align} E(g(x)) &\geq g(E(x))\\ E(f(-x)) &\geq f(-E(x))\\ E(-1/x) &\geq -1/E(x)\\ - E(-1/x) &\leq 1/E(x) \end{align}
I continue by giving a geometric argument that $-E(-1/x) = E(1/x)$. So
$$E(1/x) \leq 1/E(x).$$
I'm OK with showing the convexity and differentiability of the function, but I'm not OK with this argument, in particular this last step that I'm trying to give a geometric argument using reflections. Can you shed some light? Thanks!
It is true that if a random variable $X$ takes on positive values only then $$E\left[\frac{1}{X}\right] \ge \frac{1}{E[X]}, \tag{1}$$ and $$E\left[\frac{1}{X}\right] \le \frac{1}{E[X]}, \tag{2}$$ when $X$ takes on negative values only. But $(1)$ and $(2)$ do not imply that $$E\left[\frac{1}{X}\right] = \frac{1}{E[X]} \tag{3}$$ because $(1)$ and $(2)$ hold for different types of RVs: $(1)$ for positive and $(2)$ for negative RVs.