= Solution
Choose a <Uniformly random maximal chain in a Boolean lattice>. A fixed $h$-element set lies on the chain with probability $\binom nh^{-1}$. Since an <antichain> meets any chain at most once, the expected number of its members on the chain is at most one:
$$
\sum_{h=0}^n|\mathcal A_h|\binom nh^{-1}\leq1.
$$
This is the <Lubell--Yamamoto--Meshalkin inequality>.
For the equality case, every maximal chain must meet $\mathcal A$. Suppose $S\in\mathcal A$ has size $h$ and $T$ is obtained from $S$ by exchanging one element. There is a maximal chain whose rank-$h$ member is $T$, whose rank-$(h-1)$ member is $S\cap T$, and whose rank-$(h+1)$ member is $S\cup T$. Every lower member is contained in $S$ and every higher member contains $S$, so antichainness excludes all of them. Equality forces $T\in\mathcal A$. The graph of $h$-subsets joined by one-element exchanges is connected, hence $\mathcal A=[n]^{(h)}$. Conversely, a full level clearly gives equality.
Since $\binom nh\leq\binom n{\lfloor n/2\rfloor}$,
$$
\frac{|\mathcal A|}{\binom n{\lfloor n/2\rfloor}}
\leq\sum_h|\mathcal A_h|\binom nh^{-1}\leq1.
$$
The equality characterization above leaves exactly either middle level. This proves <Sperner theorem> with its equality cases.
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