Problem: Let $A_1,\cdots,A_m$ be sets with size $r$ and $B_1,\cdots,B_m$ be sets with size $s$. Suppose $A_j\cap B_j=\emptyset \forall j=1,\cdots,m$ and for each $j\ne k$, at least one of $A_j\cap B_k, A_k\cap B_j$ is nonempty. Show that $m\le \frac{(r+s)^{r+s}}{r^rs^s}$
My ideas: Similar to the normal Bollobas theorem, label elements in $\bigcup A_j \cup \bigcup B_j$ with distinct integers. Let $N$ be the number of $j$ such that $\max A_j < \min B_j$
Observation 1: $\mathbb{E}[N] = \frac{m}{\binom{r+s}{r}}$; this implies that proving $\mathbb{E}[N] < C \sqrt{\frac{rs}{r+s}}$ (for some constant $C$) suffices.
Observation 2: Suppose $\max A_{i_j} < \min B_{i_j}$ for $j=1,\cdots,t$. Then we can relabel $j$'s such that
$$\max A_{i_1} < \min B_{i_1} \le \max A_{i_2} < \min B_{i_2} \le \cdots \le \max A_{i_t} < \min B_{i_t}$$
Following Carl's two hints, here is a solution:
WLOG, $\bigcup A_j \cup \bigcup B_j = [k]$.
Construct a random function from $[k]$ to $[r+s]$ (finally smth that might be linked to $(r+s)^{r+s}$). Let $A_j = \{a_{j,1},\cdots,a_{j,r}\}$ and $B_j = \{b_{j,1},\cdots, b_{j,s}\}$. Let $E_j$ be the event that
$$\{ f(a_{j,t}) | 1\le t\le r \} = [r] \text{ and } \{ f(b_{j,t}) | 1\le t\le s \} = \{r+1,\cdots,r+s\}$$
Note that $\mathbb{P}(E_j) = \frac{r^rs^s}{(r+s)^{r+s}}$
The $E_j$'s are pairwise disjoint; for if $E_j,E_k$ both occur, WLOG $A_j\cap B_k \ne \emptyset$. This implies that there exists $t_1,t_2$ such that $$f^{-1}(\{1,\cdots,r\}) \ni a_{j,t_1} = b_{j,t_2} \in f^{-1}(\{r+1,\cdots,r+s\})$$
Absurd!
This implies that $$1\ge \mathbb{P}(\bigcup E_j) = \sum_{j=1}^m \mathbb{P}(E_j) = \frac{mr^rs^s}{(r+s)^{r+s}}$$
The conclusion follows.
(This can also be phrased in terms of expected value; let $X$ be the number of $E_j$'s that are true, then $$\frac{mr^rs^s}{(r+s)^{r+s}}=\mathbb{E}[X] \le 1$$)