Does $Var(X^2) \geq (VarX)^2$ hold? It is well known that $E(X^2) \geq (EX)^2$, but I was wondering if there is a similar result for variances, e.g. is $Var(X^2) \geq (VarX)^2$?
I was doing some research and came up with that inequality, but I can’t prove it. I’ve done simulations in R for several known distributions (e.g. bernoulli, binomial, poisson, normal, gamma, t, exponential, all for few parameters) and they seem to show that it really does hold, but I’m not sure whether it holds generally.
If this doesn’t hold, is there any other result that somehow links $Var(X^2)$ and $(VarX)^2$?
EDIT: So, as H. H. Rugh showed in his answer, it can't hold generally, but it often does. I'm still interested in a reference to a different result that gives a certain link between the two expressions, or perhaps some sufficient conditions for the inequality, etc.

Nice question. It can't hold in general, since you may take a r.v. with values in e.g. $\pm 1$, for which $\text{var}(X^2)=0 < \text{var}(X)^2$. 
But numerically it seems to hold when $X$ is sign-definite, e.g. $X\geq 0$ a.s. (in accordance with your simulations).
