Lets assume you have a probability distribution $X$, with mean $\mu$ and variance $\sigma^2$. Let $$\bar{\mu} = \frac{1}{n} \sum_{i=1}^n x_i$$ be the sample mean of $X$, and let $$\bar{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{\mu})^2$$ be its biased estimator (you can assume unbiased estimator too). I want to find $$\mathbb{E} \left[\frac{\bar{\mu}}{\bar{\sigma}^2}\right],$$ i.e. the expectation of sample mean to sample variance. I already know for gaussian distribution we can distribute the expectation operator, but what if $X$ is not a gaussian distribution. Is there some method or way I can distribute the expectation operator and attain a upper bound or approximate value of it i.e. $$\mathbb{E} \left[\frac{\bar{\mu}}{\bar{\sigma}^2}\right] \leq \mathbb{E}[\bar{\mu}] \mathbb{E}\left[\frac{1}{\bar{\sigma}^2}\right] + C$$ I want to find this $C$, and more specifically a way to distribute the expectation operator.
My solutions have not found anything solid and I don't think my ways are viable. Anyone who can give any hints on how to solve it would be appreciated.
EDIT: Possible Solution:
$$\mathbb{E} \left[\frac{\bar{\mu}}{\bar{\sigma}^2}\right] = \mathbb{E}[\bar{\mu}] \mathbb{E}\left[\frac{1}{\bar{\sigma}^2}\right] + Cov\left(\bar{\mu},\frac{1}{\bar{\sigma}^2}\right) \leq \mathbb{E}[\bar{\mu}] \mathbb{E}\left[\frac{1}{\bar{\sigma}^2}\right] + \sqrt{Var(\bar{\mu})Var\left(\frac{1}{\bar{\sigma}^2}\right)}$$
We know already know $\mathbb{E}[\bar{\mu}]$ and $Var(\bar{\mu})$, but how can I find $\mathbb{E}\left[\frac{1}{\bar{\sigma}^2}\right]$, and $Var\left(\frac{1}{\bar{\sigma}^2}\right)$. I have already derived a solution for $\mathbb{E}\left[\frac{1}{\bar{\sigma}^2}\right]$, but only for gaussian distribution. I am not getting any idea on how to solve $\mathbb{E}\left[\frac{1}{\bar{\sigma}^2}\right]$, and $Var\left(\frac{1}{\bar{\sigma}^2}\right)$ for any distribution