Every union closed family $\mathcal{F}$, $\emptyset \notin \mathcal{F}$, with $|\mathcal{F}| = n$ sets, must have at least $\frac{2}{3}\binom{n}{2}$ unordered couples of sets with at least one element in common. This is because for every couple without an element in common $\{A,B\}$ we can build two other couples $\{A,A \cup B\}$ and $\{B,A \cup B\}$ with non-empty intersection between the two sets, and also $\{A,A \cup B\} = \{A,A \cup C\}$ implies $B = C$ when $A \cap B = \emptyset$ and $A \cap C = \emptyset$.
The bound is tight for $n = 3$. However, is it possible to improve it for $n$ bigger than some value? For example to $\frac{3}{4}\binom{n}{2}$? Or to some $f(n)\binom{n}{2}$ with $f(n)$ increasing? Might considering only the possible counterexamples of the union closed sets conjecture help?
