Choose disjoint <subsets> $K_1,K_2\subseteq[n]$ of size $k$ and let $R=[n]\setminus(K_1\cup K_2)$. The <two-block extremisers for the cross-Sperner inequality> are
$$
\mathcal A=\{S\subseteq[n]:K_1\subseteq S,\ S\cap K_2=\varnothing\},\qquad \mathcal B=\{S\subseteq[n]:K_1\nsubseteq S,\ S\cap K_2\ne\varnothing\}.
$$
For $A\in\mathcal A$ and $B\in\mathcal B$, some point of $K_1$ belongs to $A$ but not $B$, so $A\nsubseteq B$. Some point of $K_2$ belongs to $B$ but not $A$, so $B\nsubseteq A$. Thus these form a <cross-Sperner family> pair.

An element of $\mathcal A$ fixes its intersections with both blocks and freely chooses its <subset> of $R$. An element of $\mathcal B$ chooses any proper <subset> of $K_1$, any nonempty <subset> of $K_2$, and any <subset> of $R$. Hence
$$
\boxed{|\mathcal A|=2^{n-2k}=2^{-2k}2^n,\qquad |\mathcal B|=(2^k-1)^2\,2^{n-2k}=(1-2^{-k})^2\,2^n.}
$$
The hypotheses ensure that the blocks exist and both <set families> are nonempty, including when $n=2k$. Moreover,
$$
\sqrt{|\mathcal A|}+\sqrt{|\mathcal B|}=\bigl(1+(2^k-1)\bigr)2^{(n-2k)/2}=2^{n/2},
$$
so this construction attains equality in the <Cross-Sperner inequality>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 109 / 3 / iii / Solution

Choose disjoint <subsets> $K_1,K_2\subseteq[n]$ of size $k\geq1$ with $2k\leq n$. Let $\mathcal A$ consist of <subsets> containing all of $K_1$ and none of $K_2$; let $\mathcal B$ consist of <subsets> omitting at least one point of $K_1$ and containing at least one point of $K_2$. These form a <cross-Sperner family> pair, with sizes $2^{n-2k}$ and $(2^k-1)^2 2^{n-2k}$. Their square roots sum to $2^{n/2}$, proving sharpness of the <Cross-Sperner inequality>. The free coordinates outside the two blocks account for the factor $2^{n-2k}$.

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 Incoming links: Two-block extremisers for the cross-Sperner inequality