Q:

Simple probability inequality to show

How can I show that 
$ P(A \cup B) P(A\cap B) \le P(A) P(B)$ for any events A and B?
I have tried using the inclusion/exclusion principle and using conditional probability but I keep going round in circles.
Thanks

A:

Let $X=A\backslash B$, $Y=A\cap B$, $Z=B\backslash A$ be three disjoint events and $x=P(X)$, $y=P(Y)$, $z=P(Z)$ ($x,y,z \geq 0$). 
Then:
$$P(A)=x+y\\P(B)=y+z\\P(A\cup B)=x+y+z\\P(A\cap B)=y$$
So
$$P(A)P(B)-P(A\cup B)P(A\cap B)=\\(x+y)(y+z)-y(x+y+z)=\\xy+xz+y^2+yz-xy-y^2-yz = \\ xz \geq 0$$
Thus:
$$P(A)P(B)\geq P(A\cup B)P(A\cap B)$$

