Given are three unit disks on the plane. Let $A$ be the area of the plane covered by exactly $1$ disk. Let $B$ be the area of the plane covered by exactly $2$ disks. Prove that $A\geq B$.
Intuitively, this statement seems unlikely. One might think that if one draws three disks as in the Venn diagram, with a lot of overlapping, then one can get $A<B$. But apparently a large area is covered by all three disks, which is not counted toward $A$ or $B$.