Consider a union-closed family of sets $\mathcal{F}$, with $n = \vert\mathcal{F}\vert$ and thus $n \choose 2$ unordered couples of distinct sets $\{A, B\}$, $A,B \in \mathcal{F}$.

In general, the minimum number of couples such that $A \subset B$ or $B \subset A$ is $n-1$, reached with ${\mathcal{F}} = A \;\cup\; \{B_i \;:\; i=1,\ldots,n-1\}, $ where $A = \{1,2,\ldots,n-1\}$, and $B_i = A\setminus\{i\}$ (see here).

We add the requirement that there must exist $\lceil (n+1)/2 \rceil$ sets in $\mathcal{F}$ such that their union is different from the universe $U(\mathcal{F})$ (the union of all sets in $\mathcal{F}$). This is equivalent to say that not all elements are abundant, i.e. belong to at least half of the sets in $\mathcal{F}$.

In this case, can we get a lower bound $f(n)$ better than $n-1$:

$$\text{Number of couples such that } A \subset B \text{ or } B \subset A \ge f(n) \gt n-1$$?

Any idea?