covering space by circles Let $\Lambda$ be a lattice of $\mathbb R^2$. There is a result that says
If when a circle of radius $1$ is centered at each lattice point, the circles will cover all of $\mathbb R^2$ then the area of the fundamental parallelogram $\leq \frac{3}{2}\sqrt3$
Can someone provide me the proof of this or point to a paper or textbook that proofs this result. It's apparently well known.
EDIT: Rephrasing this question:
Consider a parallelogram $P$. Suppose that the distance between any point inside the parallelogram and it's nearest vertex is at most $1$ then the area of the parallelogram $\leq \frac{3}{2}\sqrt3$
 A: Let $2d>0$ be the smallest occurring distance between two lattice points. Then $0< d\leq 1$. We may assume that $(\pm d,0)$ are lattice points and that the line $\ell: \ y=h$ is the first horizontal lattice line above $y=0$. By definition of $d$ one necessarily has $$h\geq\sqrt{3} d\ .\tag{1}$$
The point $P:=(0,\sqrt{1-d^2})$ is on the rims of both unit disks with centers $(\pm d,0)$, and these disks do not cover a full neighborhood of $P$. If $P$ had a distance $>1$ from all lattice points on $\ell$ some points in the neighborhood of $P$ would still be left uncovered. It follows that there is a lattice point $(u,h)\in\ell$ with $|u|\leq d$ and
$$|P-(u,h)|^2=u^2+\bigl(h-\sqrt{1-d^2})^2\leq1\ .$$
This implies $h\leq1+\sqrt{1-d^2}$, and together with $(1)$ we obtain that $0<d\leq{\sqrt{3}\over2}$.
It follows that the area $A$ of a fundamental parallelogram satisfies
$$A=2d\>h\leq 2d\bigl(1+\sqrt{1-d^2}\bigr)=:f(d)\ .$$
Maximizing $f$ over $0<d\leq{\sqrt{3}\over2}$ proves that indeed
$$A\leq{3\sqrt{3}\over2}\ ,$$
which corresponds to the regular hexagonal lattice, as expected.
