I came across the following inequality in one of my calculations ($X,Y$ are centered random variables):


$$\operatorname{E}(X^2Y^2)-\operatorname{E}(X^2)\operatorname{E}(Y^2) \geq 2 \operatorname{E}(XY)^2$$


or, written in terms of covariances,


$$\operatorname{Cov}(X^2,Y^2) \geq 2 \operatorname{Cov}(X,Y)^2$$.


If $(X,Y)=(U,V)$ is a two-dimensional centered Gaussian, this becomes an equality and if $(X,Y)=(H\_p(U),H\_q(V))$, where $(U,V)$ is still a two-dimensional centered Gaussian with $\operatorname{E}(U^2)=\operatorname{E}(V^2)=1$ and $H\_k$ denotes the $k$th (probabilists') Hermite polynomial, the inequality above is strict whenever $p,q \geq 2$.


I have a feeling that something like this should be true for arbitrary random variables but couldn't prove it. Does this inequality look familiar to anybody or do you have an idea on how this could be proved/disproved?



Is the covariance of squares always bounded from below by two times the covariance?
An easy counterexample: $X=Y$, $P(X=\pm1)=1/2$. Then the left side of your inequality is $0$, and its right side is $2$. 


A much more general, and perhaps more instructive, class of counterexamples is as follows. Let $U$ and $V$ be any random variables (r.v.'s) with values in $(0,\infty)$ such that $Cov(U,V)\le0$. For instance, one may 
let $V:=g(U)$, where $g\colon(0,\infty)\to(0,\infty)$ is a decreasing function, so that the r.v.'s $U$ and $V$ be negatively associated. Let now $X:=\varepsilon\sqrt U$ and $Y:=\varepsilon\sqrt V$, where $\varepsilon$ is a Rademacher r.v. (with $P(\varepsilon=\pm1)=1/2$) independent of $U$ and $V$. 
Assume that $E(\sqrt U+\sqrt V)<\infty$. 
Then $EX=EY=0$ and $XY=\sqrt{UV}>0$, so that the right side of your inequality is $>0$. On the other hand, $Cov(X^2,Y^2)=Cov(U,V)\le0$, that is, the left side of your inequality is $\le 0$, so that the inequality fails to hold. 


