TITLE: Does $Var(X^2) \geq (VarX)^2$ hold?
QUESTION [6 upvotes]: 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.

REPLY [5 votes]: Let me prove H. H. Rugh's prediction that the inequality is true if
$$ \mathsf{P}(X \geq 0) = 1 \qquad \text{and}\qquad \mathsf{E}[X^4] < \infty. $$
Let $Y \stackrel{d}{=} X$ be an independent copy of $X$. Then notice that
$$ 2\mathsf{Var}(X) = \mathsf{E}[(X - Y)^2] \qquad \text{and} \qquad 2\mathsf{Var}(X^2) = \mathsf{E}[(X^2-Y^2)^2]. $$
Now since $X+Y \geq |X-Y|$, if we set $Z = (X-Y)^2$ then
$$ 2\mathsf{Var}(X^2) = \mathsf{E}[(X-Y)^2(X+Y)^2] \geq \mathsf{E}[Z^2] \geq \mathsf{E}[Z]^2 = 4\mathsf{Var}(X)^2. $$
This actually proves a stronger inequality $ \mathsf{Var}(X^2) \geq 2\mathsf{Var}(X)^2$.

Addendum. Here is an optimal result of this kind:

Claim. If $X \geq 0$ a.s. and $\mathsf{E}[X^4] < \infty$, then we have
$$\mathsf{Var}(X^2) \geq 4 \mathsf{Var}(X)^2.$$
The equality holds if and only if $X$ is either constant or a multiple of the Bernoulli distribution of parameter $\frac{1}{2}$.

Proof. Let $\mu = \mathsf{E}X = \mathsf{E}Y$ denote the common expectation of $X$ and $Y$. Following the previous computation, we find that
\begin{align*}
\mathsf{Var}(X^2)
&= \frac{1}{2}\mathsf{E}[(X-Y)^2(X+Y)^2]
 \geq \frac{1}{2}\mathsf{E}[(X-Y)^4] \tag{1} \\
&= \frac{1}{2}\mathsf{E}[((X-\mu)-(Y-\mu))^4]
 = \mathsf{E}[(X-\mu)^4] + 3\mathsf{Var}(X)^2 \\
&\geq 4\mathsf{Var}(X)^2. \tag{2}
\end{align*}
In order for this to be an equality, we need that $\text{(1)}$ and $\text{(2)}$ becomes equality. To avoid unnecessary complication, assume WLOG that $X$ is non-constant.
At $\text{(1)}$, we must have $(X+Y)^2 = (X-Y)^2$ whenever $X \neq Y$. Equivalently, $XY = 0$ must hold whenever $X \neq Y$. This forces that there are at most one non-zero value that $X$ can attain. Indeed, if there are two disjoint Borel sets $B_1, B_2 \subseteq (0, \infty)$ such that $\mathsf{P}(X \in B_i) > 0$ for $i = 1, 2$, then we must have
$$0 = \mathsf{P}(XY \neq 0, X\neq Y) \geq \mathsf{P}(X \in B_1)\mathsf{P}(Y \in B_2) > 0, $$
a contradiction. So there exists $c > 0$ such that $\mathsf{P}(X = c) = p \in (0, 1)$ and $\mathsf{P}(X = 0) = 1-p$. Then
$$ \mathsf{Var}(X^2) = c^4 p(1-p) \qquad \text{and} \qquad 4\mathsf{Var}(X)^2 = 4c^4 p^2(1-p)^2. $$
For these to coincide, we must have $p = \frac{1}{2}$. ////