# Lower bound for sets couples such that $A \subset B$ or $B \subset A$ in some union-closed families

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?

For $$n\ge3$$ let $$f(n)$$ denote the minimum possible number of $$2$$-element chains in a union-closed family $$\mathcal F$$ of cardinality $$n$$ in which not every element of $$\bigcup\mathcal F$$ is abundant.
The obvious lower bound $$f(n)\ge n-1+\left\lceil\frac{n+1}2\right\rceil-1=\left\lceil\frac{3(n-1)}2\right\rceil$$ is attained by the family consisting of the $$n=h+k+2$$ sets
$$S\setminus\{x_0,x_1\},\,S\setminus\{x_0,x_2\},\dots,S\setminus\{x_0,x_h\},\,S\setminus\{x_0\},$$
$$S\setminus\{y_1\},\,S\setminus\{y_2\},\dots,S\setminus\{y_k\},\,S$$
where $$h=\left\lceil\frac{n-1}2\right\rceil$$, $$k=\left\lfloor\frac{n-3}2\right\rfloor$$, $$|S|=h+k+1=n-1$$, and $$S=\{x_0,x_1,\dots,x_h,y_1,\dots,y_k\}.$$
Hence $$f(n)=\left\lceil\frac{3(n-1)}2\right\rceil.$$