Define
\begin{equation*}
M=\binom{n}{\lfloor n/2\rfloor}
\qquad\text{and}\qquad
L=\left\lceil\frac{n+1}{2}\right\rceil.
\end{equation*}
First suppose that \(\emptyset\notin\mathcal{F}\text{.}\) Let \(\mathcal{P}=2^{[n]}\setminus\{\emptyset\}\text{.}\) Choose a uniformly random maximal chain in \(2^{[n]}\) and delete the empty set from it; call the resulting random chain \(\mathcal{C}\) in \(\mathcal{P}\text{.}\) For every \(A\in\mathcal{P}\text{,}\)
\begin{equation*}
\mathbb{P}(A\in\mathcal{C})=\binom{n}{|A|}^{-1}\geq 1/M.
\end{equation*}
We obtain a comparability digraph
\(D\) for
\(\mathcal{P}\) by orienting comparable pairs as follows. If
\(A\) and
\(B\) are comparable and
\(|A|\neq |B|\text{,}\) orient the pair from the set whose cardinality is farther from
\(n/2\) to the set whose cardinality is closer to
\(n/2\text{.}\) In the case of a tie, so that
\(A\subsetneq B\) and
\(|A|=n-|B|\text{,}\) orient the pair from
\(A\) to
\(B\text{.}\)
We claim that, for every arc \((X,Y)\) in \(D\text{,}\)
\begin{equation*}
\mathbb{P}(Y\in\mathcal{C}\mid X\in\mathcal{C})\leq 1/L.
\end{equation*}
Suppose first that \(X\subsetneq Y\text{.}\) Then
\begin{equation*}
\mathbb{P}(Y\in\mathcal{C}\mid X\in\mathcal{C})
=
\binom{n-|X|}{|Y|-|X|}^{-1}.
\end{equation*}
Since the arc is oriented from \(X\) to \(Y\text{,}\) the set \(X\) is at least as far from the middle level as \(Y\text{.}\) Thus \(|X|<n/2\text{.}\) Also, since \(X\neq\emptyset\text{,}\) the endpoint \(Y\) cannot be all of \([n]\text{;}\) otherwise \(Y\) would be farther from the middle than \(X\text{.}\) Hence
\begin{equation*}
n-|X|\geq L
\qquad\text{and}\qquad
1\leq |Y|-|X|<n-|X|.
\end{equation*}
Therefore
\begin{equation*}
\binom{n-|X|}{|Y|-|X|}\geq L.
\end{equation*}
So
\begin{equation*}
\mathbb{P}(Y\in\mathcal{C}\mid X\in\mathcal{C})\leq 1/L.
\end{equation*}
The case \(Y\subsetneq X\) is similar. In this case,
\begin{equation*}
\mathbb{P}(Y\in\mathcal{C}\mid X\in\mathcal{C})
=
\binom{|X|}{|X|-|Y|}^{-1}.
\end{equation*}
Since the arc is oriented from \(X\) to \(Y\text{,}\) the set \(X\) is at least as far from the middle level as \(Y\text{.}\) Thus \(|X|>n/2\text{.}\) Also \(Y\neq\emptyset\text{,}\) since \(\emptyset\notin\mathcal{P}\text{.}\) Therefore
\begin{equation*}
|X|\geq L
\qquad\text{and}\qquad
1\leq |X|-|Y|<|X|.
\end{equation*}
It follows that
\begin{equation*}
\binom{|X|}{|X|-|Y|}\geq L,
\end{equation*}
as required.
\begin{equation*}
|\{(A,B)\in\mathcal{F}^2:A\subsetneq B\}|
\geq
L(|\mathcal{F}|-M).
\end{equation*}
This proves the result when \(\emptyset\notin\mathcal{F}\text{.}\)
Now suppose that \(\emptyset\in\mathcal{F}\text{.}\) Apply the previous case to \(\mathcal{F}'=\mathcal{F}\setminus\{\emptyset\}\text{.}\) The empty set is comparable to every other set in \(\mathcal{F}\text{,}\) so
\begin{equation*}
\begin{aligned}
|\{(A,B)\in\mathcal{F}^2:A\subsetneq B\}|
&\geq
L\left(|\mathcal{F}'|-M\right)+|\mathcal{F}'|\\
&=
L\left(|\mathcal{F}|-M\right)+|\mathcal{F}'|-L.
\end{aligned}
\end{equation*}
If \(|\mathcal{F}'|\geq L\text{,}\) this is enough. If \(|\mathcal{F}'|<L\text{,}\) then \(|\mathcal{F}|\leq L\text{.}\) Since \(L\leq M\text{,}\) the quantity \(L(|\mathcal{F}|-M)\) is non-positive, and the desired inequality is trivial.
