Let $X$ be a random variable with finite support contained in $(0,\infty)$. By Jensen's inequality we have $\mathbb{E}[1/X] \geq 1/\mathbb{E}[X]$. I am curious if there exists an elementary proof of this fact which avoids the use of Jensen's inequality.
To rephrase the setup in an elementary way:
Let $S$ be a finite set of positive real numbers, and let $f : S \to [0,1]$ be a function such that $\sum_{s \in S} f(s) = 1$. Then we want to show that
$$\sum_{s \in S} f(s)/s \geq \left(\sum_{s \in S} f(s) s \right)^{-1}.$$
Equivalently, we want to show that
$$\sum_{s, s' \in S} f(s)f(s') \frac{s}{s'} \geq 1.$$
What is the most elementary proof of this fact? In particular, is there a nice proof which avoids arguing by induction? I would like to present a proof of this to students who are good at algebraic manipulation but who haven't learned Jensen's inequality and are not so comfortable with writing proofs by induction.
Here's something that might be useful:
$$\operatorname{Cov}(X,1/X) = \mathbb{E}[X(1/X)] - \mathbb{E}[X]\mathbb{E}[1/X] = 1-\mathbb{E}[X]\mathbb{E}[1/X],$$
We should expect that $X$ and $1/X$ are nonpositively correlated, because when $X$ is relatively large $1/X$ is relatively small, and visa-versa. If we can prove this, then we have $1 - \mathbb{E}[X]\mathbb{E}[1/X] \leq 0$, and we are done.
It can be proved by exploiting the monotonicity of the function $x \mapsto 1/x$ on $(0, \infty)$. Let $X_1, X_2 \text{ i.i.d. } \sim X$. It is easy to verify by monotonicity that \begin{align*} (X_1 - X_2)\left(\frac{1}{X_1} - \frac{1}{X_2}\right) \leq 0. \end{align*} Taking expectations on both sides and using the i.i.d. of $X_1, X_2$ then yields \begin{align} 1 - E[X_1]E\left[\frac{1}{X_2}\right] - E[X_2]E\left[\frac{1}{X_1}\right] + 1 = 2 - 2E[X]E\left[\frac{1}{X}\right]\leq 0, \end{align} which is the desired inequality.