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If $\mathcal{A}$ is an antichain in $\mathcal{P}([n])$ and ${\mathcal{A}}_h$ is its rank-$h$ part, thenThe left side is the expected number of members of $\mathcal{A}$ on a Uniformly random maximal chain in a Boolean lattice.

For $\mathcal{A}\subseteq [n{]}^{(r)}$, the Local LYM inequality isCount containment pairs $(B,A)$ with $B\in \partial \mathcal{A}$, $A\in \mathcal{A}$, and $B\subset A$. Every $A$ contributes $r$ pairs, while every $B$ contributes at most $n-r+1$. Hencewhich rearranges to the claimed normalized inequality.

The LYM inequality says that every antichain $\mathcal{F}\subseteq \mathcal{P}([n])$ satisfiesFor the local-LYM proof, replace the highest nonempty level of $\mathcal{F}$ by its lower shadow. No shadow member contains a surviving member, since that would make the original family non-antichain, and the local inequality says that the normalized size does not decrease. Repeating pushes the family to the bottom level, whose normalized size is at most one. Therefore the original normalized sum was at most one.

For the chain proof, choose a Uniformly random maximal chain in a Boolean lattice. A fixed $r$-set belongs to it with probability ${\left(\genfrac{}{}{0ex}{}{n}{r}\right)}^{-1}$. Since an antichain meets each chain at most once, the expected number of its members on the chain is at most one. Linearity of expectation gives the displayed sum.

Finally use a symmetric chain decomposition of a Boolean lattice. Convexity makes the intersection of $\mathcal{A}$ with each symmetric chain an interval. The alternating sum of $(-1{)}^{|A|}$ over an interval in a chain is $0$, $1$, or $-1$. There are exactly $\left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$ chains, because each contains exactly one member of a middle level. Summing the chain contributions gives

Choose a Uniformly random maximal chain in a Boolean lattice. A fixed $h$-element set lies on the chain with probability ${\left(\genfrac{}{}{0ex}{}{n}{h}\right)}^{-1}$. Since an antichain meets any chain at most once, the expected number of its members on the chain is at most one: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 $\left(\genfrac{}{}{0ex}{}{n}{h}\right)\le \left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$,The equality characterization above leaves exactly either middle level. This proves Sperner theorem with its equality cases.