[proofplan]
We prove the lower bound by exhibiting the middle rank of the Boolean lattice as an antichain. For the upper bound, we take an arbitrary antichain and apply the [Lubell-Yamamoto-Meshalkin inequality](/theorems/8091) for Boolean lattices. The remaining ingredient is the elementary fact that the binomial coefficients are largest at the middle rank, which turns the LYM sum into a direct bound on the size of the antichain.
[/proofplan]
[step:Exhibit the middle rank as an antichain]
Let $m:=\lfloor n/2\rfloor$, and define
\begin{align*}
\mathcal M:=\{S\subseteq [n]: |S|=m\}.
\end{align*}
If $S,T\in \mathcal M$ and $S\subseteq T$, then $|S|=|T|$, so $S=T$. Thus no two distinct elements of $\mathcal M$ are comparable in $B_n$, and $\mathcal M$ is an antichain. Its cardinality is
\begin{align*}
|\mathcal M|=\binom{n}{m}.
\end{align*}
Since $w(B_n)$ is the maximum cardinality of an antichain in $B_n$, this gives
\begin{align*}
w(B_n)\ge \binom{n}{\lfloor n/2\rfloor}.
\end{align*}
[/step]
[step:Show the middle binomial coefficient dominates every rank size]
We claim that for every integer $k$ with $0\le k\le n$,
\begin{align*}
\binom{n}{k}\le \binom{n}{m}.
\end{align*}
For $0\le k<m$, the ratio of consecutive binomial coefficients is
\begin{align*}
\frac{\binom{n}{k+1}}{\binom{n}{k}}=\frac{n-k}{k+1}.
\end{align*}
Since $k<m\le n/2$, we have $n-k\ge k+1$, so this ratio is at least $1$. Hence the sequence $\binom{n}{0},\binom{n}{1},\dots,\binom{n}{m}$ is nondecreasing. For $k>m$, use the symmetry
\begin{align*}
\binom{n}{k}=\binom{n}{n-k}.
\end{align*}
The integer $n-k$ lies below or at the opposite middle rank, so the same monotonicity up to the middle gives
\begin{align*}
\binom{n}{n-k}\le \binom{n}{m}.
\end{align*}
Therefore $\binom{n}{k}\le \binom{n}{m}$ for every $0\le k\le n$.
[/step]
[step:Apply the LYM inequality to bound an arbitrary antichain]
Let $A\subseteq B_n$ be an arbitrary antichain. By the Lubell-Yamamoto-Meshalkin inequality for Boolean lattices [citetheorem:8091], applied to the antichain $A$ in the Boolean lattice $B_n$, we have
\begin{align*}
\sum_{S\in A}\frac{1}{\binom{n}{|S|}}\le 1.
\end{align*}
For each $S\in A$, the previous step gives
\begin{align*}
\binom{n}{|S|}\le \binom{n}{m}.
\end{align*}
Taking reciprocals of these positive integers yields
\begin{align*}
\frac{1}{\binom{n}{|S|}}\ge \frac{1}{\binom{n}{m}}.
\end{align*}
Summing this inequality over all $S\in A$ gives
\begin{align*}
\sum_{S\in A}\frac{1}{\binom{n}{|S|}}\ge \sum_{S\in A}\frac{1}{\binom{n}{m}}=\frac{|A|}{\binom{n}{m}}.
\end{align*}
Combining the two inequalities gives
\begin{align*}
\frac{|A|}{\binom{n}{m}}\le 1.
\end{align*}
Thus
\begin{align*}
|A|\le \binom{n}{m}.
\end{align*}
[guided]
Let $A\subseteq B_n$ be any antichain. We want to show that $A$ cannot have more elements than the middle rank. The tool that measures an antichain across all ranks at once is the Lubell-Yamamoto-Meshalkin inequality for Boolean lattices [citetheorem:8091]. Its hypothesis is exactly that $A$ is an antichain in the Boolean lattice $B_n$, so it applies and gives
\begin{align*}
\sum_{S\in A}\frac{1}{\binom{n}{|S|}}\le 1.
\end{align*}
The denominator $\binom{n}{|S|}$ is the size of the rank containing $S$. The previous step showed that every rank has size at most the middle rank size:
\begin{align*}
\binom{n}{|S|}\le \binom{n}{m}
\end{align*}
for every $S\in A$, where $m=\lfloor n/2\rfloor$. Since both binomial coefficients are positive, taking reciprocals reverses the inequality:
\begin{align*}
\frac{1}{\binom{n}{|S|}}\ge \frac{1}{\binom{n}{m}}.
\end{align*}
Now sum this lower bound over all elements of the antichain $A$. The right-hand side is constant in $S$, so
\begin{align*}
\sum_{S\in A}\frac{1}{\binom{n}{|S|}}\ge \sum_{S\in A}\frac{1}{\binom{n}{m}}=\frac{|A|}{\binom{n}{m}}.
\end{align*}
Together with the [LYM inequality](/theorems/2586), this gives
\begin{align*}
\frac{|A|}{\binom{n}{m}}\le 1.
\end{align*}
Multiplying by the positive integer $\binom{n}{m}$ yields
\begin{align*}
|A|\le \binom{n}{m}.
\end{align*}
Thus every antichain in $B_n$ has cardinality at most the size of the middle rank.
[/guided]
[/step]
[step:Take the maximum over all antichains]
The preceding step proves that every antichain $A\subseteq B_n$ satisfies
\begin{align*}
|A|\le \binom{n}{m}.
\end{align*}
Therefore
\begin{align*}
w(B_n)\le \binom{n}{m}.
\end{align*}
The first step gave the reverse inequality. Hence
\begin{align*}
w(B_n)=\binom{n}{m}=\binom{n}{\lfloor n/2\rfloor}.
\end{align*}
This proves the theorem.
[/step]
