Let $\left \{ \left ( A_{i},B_{i} \right ),1\leq i\leq h \right \}$ be a family of pairs of subsets of the set of integers such that $\left | A_{i} \right |=k$ for all $ i$ and $\left | B_{i} \right |=l$ for all $i$, $A_{i}\cap B_{i}=\emptyset$ and $(A_{i}\cap B_{j})\cup (A_{j}\cap B_{i})\neq \emptyset$ for all $i\neq j$ . Prove that $$h< \left(\dfrac{k+l}{k}\right)^k\left(\dfrac{k+l}{l}\right)^l.$$ [Source: The probabilistic method, Alon and Spencer, p.11]
The idea should be to set up an event with probability $\left(\dfrac{k}{k+l}\right)^k\left(\dfrac{l}{k+l}\right)^l$, but it's not clear how to do this.
(This question was posted here, but the answer posted is not understandable.)
Let $p=\frac{k}{k+l},q=1-p=\frac{l}{k+l}$.
Let $Z$ be the finite set $\cup_{k}A_k\cup B_k$. Consider a random experiment $E$ whose probability of success is $p$. Repeat this experiment independently as many times as there are elements in $Z$ ; this produces experiments $(E_{z})_{z\in Z}$.
Next, for $1\leq i \leq h$ consider the event $C_i$ : “$E_z$ succeeds for every $z\in A_i$ and fails for every $z\in B_i$”.
I claim that the $C_i$ are mutually incompatible. Indeed, let $i\neq j$ be two indices in $[1,h]$. By hypothesis, we have for example $A_i\cap B_j \neq \emptyset$ ; let $t\in A_i\cap B_j$. Then $E_t$ succeeds if $C_i$ holds, and fails if $C_j$ holds. This shows my claim. It follows that $P(\cup_{i=1}^{h}C_i)=\sum_{i=1}^h P(C_i)$. Now, the union $\cup_{i=1}^{h}C_i$ does not exhaust the whole probabilistic universe we are considering (consider for example the case when every $E_z$ succeeds). It follows that $\sum_{i=1}^h P(C_i)<1$. Now each $C_i$ occurs with probability $p^kq^l$, so $hp^kq^l <1$ as wished.