Can the variance of the sample variance be negative?

Question

In this answer is shown that the variance of the sample variance is

$$ \text{Var}(S^2) = \frac{1}{n} \left(\mu_4 - \frac{n-3}{n-1}\sigma^4\right) $$

where $\mu_4$ is the fourth central moment, ie $E[(X-\mu)^4]$.

My question is, what prevents the variance from being negative? As far as I know, it can happen that $\mu_4 < \frac{n-3}{n-1}\sigma^4$, and then the variance would be negative, which doesn't make sense.

Am I missing something?

Answer 1

By Cauchy-Schwarz,

$$\frac{n-3}{n-1}\sigma^4\leq \sigma^4=(E[(X-\mu)^2])^2\leq E[(X-\mu)^4]=\mu_4. $$