[proofplan] We count maximal chains in the Boolean lattice in two ways. First, maximal chains are identified with permutations of $[n]$, so there are $n!$ of them. Then we count the pairs $(S,C)$ where $S\in A$ and $C$ is a maximal chain containing $S$. A fixed set $S$ of size $k$ lies in exactly $k!(n-k)!$ maximal chains, while the antichain condition implies that each maximal chain contains at most one element of $A$. [/proofplan]

[step:Identify maximal chains with permutations of $[n]$] Let $\mathfrak C$ denote the set of maximal chains in $B_n$. For a permutation $\pi=(i_1,\dots,i_n)$ of $[n]$, define the chain \begin{align*} C_\pi:=\{\varnothing,\{i_1\},\{i_1,i_2\},\dots,\{i_1,\dots,i_n\}\}\subset B_n. \end{align*} This is a maximal chain because it starts at $\varnothing$, ends at $[n]$, and increases by exactly one element at each step. Conversely, let $C\in \mathfrak C$. Since $C$ is maximal under inclusion among chains in $B_n$, it must contain $\varnothing$ and $[n]$; otherwise one could add one of them to $C$ and obtain a strictly larger chain. Also, if $X\subsetneq Y$ are consecutive elements of $C$ and $|Y\setminus X|\ge 2$, choose $a\in Y\setminus X$ and insert $X\cup\{a\}$ strictly between $X$ and $Y$, contradicting maximality. Hence every consecutive inclusion in $C$ increases cardinality by one. Therefore $C$ has the form \begin{align*} \varnothing=S_0\subsetneq S_1\subsetneq \cdots \subsetneq S_n=[n], \end{align*} where $|S_j|=j$ for every $j\in\{0,\dots,n\}$. For each $j\in\{1,\dots,n\}$, the set $S_j\setminus S_{j-1}$ has exactly one element; call it $i_j$. Then $(i_1,\dots,i_n)$ is a permutation of $[n]$, and $C=C_\pi$. Thus the map $\pi\mapsto C_\pi$ is a bijection from the set of permutations of $[n]$ to $\mathfrak C$. Hence \begin{align*} |\mathfrak C|=n!. \end{align*}[/step]

[guided]The reason to introduce permutations is that they give a concrete way to list every maximal chain. A permutation $\pi=(i_1,\dots,i_n)$ records the order in which the elements of $[n]$ are inserted. From it we form \begin{align*} C_\pi:=\{\varnothing,\{i_1\},\{i_1,i_2\},\dots,\{i_1,\dots,i_n\}\}\subset B_n. \end{align*} This is a chain because each set is contained in the next one. It is maximal because it already contains one set of each possible cardinality from $0$ to $n$, and a strictly increasing chain in $B_n$ can increase cardinality at least by one at every strict inclusion. Now let $C$ be an arbitrary maximal chain in $B_n$. Maximality forces $C$ to contain $\varnothing$ and $[n]$: if either endpoint were missing, it could be added while preserving the chain property. Next, suppose $X\subsetneq Y$ are consecutive elements of $C$. If $Y\setminus X$ had at least two elements, then choosing $a\in Y\setminus X$ would give a subset $X\cup\{a\}$ satisfying \begin{align*} X\subsetneq X\cup\{a\}\subsetneq Y. \end{align*} This would allow us to insert another element into $C$, contradicting maximality. Hence consecutive elements differ by exactly one inserted element. It follows that $C$ has the form \begin{align*} \varnothing=S_0\subsetneq S_1\subsetneq \cdots \subsetneq S_n=[n], \end{align*} with $|S_j|=j$. For each $j\in\{1,\dots,n\}$, define $i_j$ to be the unique element of $S_j\setminus S_{j-1}$. The sequence $(i_1,\dots,i_n)$ contains every element of $[n]$ exactly once, so it is a permutation, and the construction recovers $C$ as $C_\pi$. Thus maximal chains are in bijection with permutations of $[n]$, and there are $n!$ of them.[/guided]

[step:Count maximal chains passing through a fixed subset] Fix $S\subset [n]$, and set $k:=|S|$. A maximal chain $C_\pi$ contains $S$ if and only if the first $k$ entries of the permutation $\pi$ are exactly the elements of $S$. There are $k!$ ways to order the elements of $S$ in the first $k$ positions, and there are $(n-k)!$ ways to order the elements of $[n]\setminus S$ in the remaining positions. Therefore the number of maximal chains containing $S$ is \begin{align*} k!(n-k)!. \end{align*} [/step]

[step:Double count pairs of antichain elements and maximal chains] Define \begin{align*} \mathcal P:=\{(S,C)\in A\times \mathfrak C:S\in C\}. \end{align*} Counting $\mathcal P$ by first choosing $S\in A$, and using the preceding step with $k=|S|$, gives \begin{align*} |\mathcal P|=\sum_{S\in A}|S|!(n-|S|)!. \end{align*} On the other hand, fix a maximal chain $C\in\mathfrak C$. Since $C$ is a chain, any two distinct elements of $C$ are comparable by inclusion. Since $A$ is an antichain, $A\cap C$ contains at most one element. Hence each $C\in\mathfrak C$ contributes at most one pair to $\mathcal P$, so \begin{align*} |\mathcal P|\le |\mathfrak C|=n!. \end{align*} Combining the two counts yields \begin{align*} \sum_{S\in A}|S|!(n-|S|)!\le n!. \end{align*} [/step]

[step:Divide by the number of maximal chains and rewrite the coefficient] For every $S\in A$, the factorial identity \begin{align*} \binom{n}{|S|}=\frac{n!}{|S|!(n-|S|)!} \end{align*} gives \begin{align*} \frac{|S|!(n-|S|)!}{n!}=\frac{1}{\binom{n}{|S|}}. \end{align*} Dividing the inequality \begin{align*} \sum_{S\in A}|S|!(n-|S|)!\le n! \end{align*} by $n!$ therefore gives \begin{align*} \sum_{S\in A}\frac{1}{\binom{n}{|S|}}\le 1. \end{align*} This is the Lubell-Yamamoto-Meshalkin inequality. [/step]