Let $U(n)$ denote the maximum possible number of pairs of points in an $n$-point subset of $\Bbb R^2$ that are unit distance apart.
Let $g(n)$ be the minimum number of distinct distances determined by $n$ points in $\Bbb R^2$, that is: $$g(n)=\text{min}_{P\subset \Bbb R^2,|P|=n}\left|\{\text{dist}(x,y)\mid x,y\in P\}\right|.$$
My textbook says "Clearly $g(n)\geq {n\choose 2} / U(n)$". Why is this so? I have no idea how to even show that this is reasonable. My first thought is that there are clearly ${n\choose 2}$ pairs and if $U(n)={n\choose 2}$, i.e. they are all unit distance apart, then indeed $g(n)=1$. Otherwise as $U(n)$ gets smaller, $g(n)$ should get bigger, which seems okay intuitively. But I don't see how to show this
Actually, a thought after typing this out: Perhaps they are thinking about taking all pairs, and breaking them up into $U(n)$ sized groups, of which would would have ${n\choose 2}/U(n)$, and one cannot do better than this.
Given an $n$-set $S\subset \mathbb R^2$, consider
$$\mathscr S = \{(x,y)\,|\, x, y \in S \,\text{ and }\, x\neq y\}.$$
Obviously, $|\mathscr S| = \binom{n}2$. Now, define an equivalence relationship $\sim$ on $\mathscr S$ by
$$ (x_0,y_0) \sim (x_1,y_1) \iff d(x_0,y_0) = d(x_1,y_1).$$
The relation $\sim$ partitions $\mathscr S$ into some number $k$ of classes, each of which has most $U(n)$ elements. Hence, $k \geq \left\lceil\binom{n}2/U(n)\right\rceil \geq \binom{n}2/U(n)$.
Now, we know there are least $g(n)$ classes. Suppose that we chose $S$ a priori such that $k = g(n)$. This is possible, for otherwise $g(n)$ would not be a minimum.
With this choice, the inequality follows. Notice that the inequality depends on $n$, but not on $S$. Using our freedom to choose $S$ can make it easier to prove the inequality.