Proof of density inequality I was wondering if there is an easy way to find sufficient conditions for the following inequality to hold
$$
\int f(x,y)^2 \:\mathrm{d}x \:\mathrm{d}y - \int f(x)^2 f(y)^2 \:\mathrm{d}x\:\mathrm{d}y \geq 0,
$$
where $f$ is the density function. Do you have any suggestions?
 A: For the case of continuous r.v.'s, we are asking under which conditions
$$\int\int f(x,y)^2 \:\mathrm{d}x \:\mathrm{d}y - \int\int f(x)^2 f(y)^2 \:\mathrm{d}x\:\mathrm{d}y \;\;?\geq ? \;\;0$$
Obviously, it holds trivially as an equality if the r.v.'s are independent. If they are not independent we ask, using $f(x,y) = f(x\mid y)f(y)$,
$$ \int\int f(x\mid y)^2f(y)^2 \:\mathrm{d}x\:\mathrm{d}y  - \int\int f(x)^2 f(y)^2 \:\mathrm{d}x\:\mathrm{d}y \;\;?\geq ? \;\;0$$
$$\Rightarrow \int\int [f(x\mid y)-f(x)][f(x\mid y)+f(x)]f(y)^2 \:\mathrm{d}x\:\mathrm{d}y \geq 0$$
All these are densities, and so non-negative. This means that a sufficient (although not necessary) condition is that 
$$ f(x\mid y)-f(x) >0$$ for the whole support,
while a necessary (but not sufficient) condition is that this inequality holds for at least some of the support (the inequality is strict because we are in the sub-case of dependent r.v.s).  
This sufficient condition can be written as 
$$ f(x, y) > f(x)f(y)$$
for the whole support. But if it should hold for the whole support, then we should also have
$$\int \int f(x, y)\mathrm{d}x\:\mathrm{d}y > \int \int f(x)f(y)\mathrm{d}x\:\mathrm{d}y$$
$$\Rightarrow 1 >1 $$
which is impossible. So at least at such level of generality, it appears that no sufficient conditions may be stated.
