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In this post it is stated that due to Jensen's inequality the expected value of the reciprocal of a strictly postive random variable $X$ will satisfy:
$$\mathbb{E}\left[\frac{1}{X}\right] \geq \frac{1}{\mathbb{E}[X]}$$
My question is whether a similar inequality exists for the variance of $1/X$?
Here is an answer (to a related question) which provides a valid generalization of $\mathbb{E}\left[\frac{1}{X}\right] \geq \frac{1}{\mathbb{E}[X]}$.
What is the relation between $\mathbb{E}X^r$ and $\mathbb{[E}X]^r$, for all possible values of $r$, when $X$ is a positive random variable? This can be answered by applying Jensens's inequality, based on the convexity or concavity of $x^r$ as a function of $x$, depending on the value of $r$.
The following presumes the relevant moments exist.
$\mathbb{E}X^r \ge \mathbb{[E}X]^r$, for $r \ge 1$ or $r \le 0$ ($x^r$ is convex in both cases)
$\mathbb{E}X^r \le \mathbb{[E}X]^r$, for $0 \le r \le 1$ ($x^r$ is concave, which includes the frequently used square root)
Note that equality holds for $r = 1$, which says $\mathbb{E}X = \mathbb{E}X$, and for $r = 0$, which says $1 = 1$. $\mathbb{E}\left[\frac{1}{X}\right] \geq \frac{1}{\mathbb{E}[X]}$ of course corresponds to $r = -1$.
Going back to the OP's original question, as shown by @kjetil b halvorsen , the analog for variance does not hold. But presuming the moments exist, we see by applying the above results with $r= -2$, that $\mathbb{E}\left[\frac{1}{X^2}\right] \geq [\frac{1}{\mathbb{E}X}]^2$
First, @Dilip Sarwate comments, for such ratio variables mean and variance often do not exist and then there is little to expect. For a detailed discussion of this see I've heard that ratios or inverses of random variables often are problematic, in not having expectations. Why is that?. But, let us assume a case where expectation and variance exists.
Your question is My question is whether a similar inequality exists for the variance of $1/X$? It is not totally clear what you mean with similar but let us take it literally, that is, is it true that $$\DeclareMathOperator{\V}{\mathbb{V}} \V(\frac1{X})\cdot \V X \ge 1 \quad \text{?} $$ In that form it is clearly false, for instance take $X$ to have a uniform distribution on a very short interval close to 1, like $[0.9, 1.1]$. For that case the product of the two variances will be 0.1127601010e-4 falsifying the inequality. But maybe you where thinking of some other generalization?
