China Team Selection Test, 2022

Proof: Decompose $D$ into the following disjoint unions $D = X \sqcup Y \sqcup Z \sqcup W$, where
$$X = \{x \in D : \exists a | x, \exists b | x\}, \quad Y = \{x \in D : \exists a | x, \nexists b | x\},$$
$$Z = \{x \in D : \nexists a | x, \exists b | x\}, \quad W = \{x \in D : \nexists a | x, \nexists b | x\}.$$
Note that the assumption of the problem indicates that $A \subseteq Y, B \subseteq Z$. It suffices to prove a stronger statement: for any two nonempty subsets $A, B$ of $D$, we always have $\sqrt{|Y|} + \sqrt{|Z|} \le \sqrt{|D|}$. This inequality is equivalent to
$$|Y| + |Z| + 2\sqrt{|Y| \cdot |Z|} \le |D| = |X| + |Y| + |Z| + |W| \iff 2\sqrt{|Y| \cdot |Z|} \le |X| + |W|,$$
This inequality is implied by $|Y| \cdot |Z| \le |X| \cdot |W|$, which can be further rewritten as
$$(|X|+|Y|)(|X|+|Z|) = |X|(|X|+|Y|+|Z|)+|Y|\cdot|Z| \le |X|(|X|+|Y|+|Z|)+|X|\cdot|W| = |X|\cdot|D|.$$
Let $U = X \cup Y$ and $V = X \cup Z$. Then the above inequality becomes $|U| \cdot |V| \le |U \cap V| \cdot |D|$.
Note that $U = \{x \in D : \exists a | x\}$ satisfies: if $x \in U$ and $x | x'$, then $x' \in U$. Call such subsets of $D$ upward-closed. Similarly, $V = \{x \in D : \exists b | x\}$ is also an upward-closed subset of $D$.
Next, we prove: for any two nonempty upward-closed subsets $U, V$ of $D$, we have $|U| \cdot |V| \le |U \cap V| \cdot |D|$.
Let $n = p_1^{\alpha_1} \cdots p_k^{\alpha_k}$ be the prime factorization of $n$, and we make an induction on $k$. Write $p$ and $\alpha$ for $p_k$ and $\alpha_k$, respectively, for simplicity. Set $n = p^{\alpha}n'$. Define $D_k = \{x \in D | v_p(x) = k\}$, $U_k = U \cap D_k$, and $V_k = V \cap D_k$. For every $k = 0, 1, \dots, \alpha - 1$, for any $x \in U_k$, the upward-closure property of $U$ implies that $px \in U_{k+1}$. This means that $|U_k| \le |U_{k+1}|$, i.e. $\{|U_k|\}_k$ is increasing. Similarly, $\{|V_k|\}_k$ is increasing. Note that $\frac{1}{p^k}U_k$ and $\frac{1}{p^k}V_k$ are upward-closed subsets of $\frac{1}{p^k}D_k = D(n')$. But inductive hypothesis, we have
$$|(\frac{1}{p^k}U_k) \cap (\frac{1}{p^k}V_k)| \ge \frac{1}{|D(n')|} \cdot |(\frac{1}{p^k}U_k)| \cdot |(\frac{1}{p^k}V_k)|,$$
So we have $|U_k \cap V_k| \ge \frac{1+\alpha}{|D|} |U_k| \cdot |V_k|$. Using rearrangement inequality, we get
$$\begin{aligned} |U \cap V| &= \sum_{k=0}^{\alpha} |U_k \cap V_k| \ge \frac{1+\alpha}{|D|} \sum_{k=0}^{\alpha} |U_k| \cdot |V_k| \\ &\ge \frac{1+\alpha}{|D|} \cdot \frac{1}{1+\alpha} \left( \sum_{k=0}^{\alpha} |U_k| \right) \cdot \left( \sum_{k=0}^{\alpha} |V_k| \right) \\ &= \frac{1}{|D|} |U| \cdot |V|. \end{aligned}$$