TITLE: Dividing books between two couples
QUESTION [14 upvotes]: Two couples of boys and girls, $(b_1,g_1)$ and $(b_2,g_2)$, are dividing a pile of books. Every book will go to one of the couples, and they'll read it together. Each person has a (nonnegative) value for each book they read and a total value of $1$ for the whole set of books in the pile; moreover some subset of books gives him/her value exactly $0.5$. 
Does there exist a constant $c>0$ for which it is always possible to give everyone value at least $c$, no matter the values and the number of books?
Without the $0.5$ condition, the answer is no. For example, there is only one book that everyone values $1$, so one couple will get $0$.

REPLY [5 votes]: EDIT: This answer is wrong, except for the Claim and its proof.
Claim:
Fix a pair $p,q$ of two persons. Then there exist two disjoint sets $A,B$ of books such that $A$ has weight at least 0.25 for both $p$ and $q$ and $B$ has weight at least 0.5 for both $p$ and $q$.
As a shorthand, I say that $A$ has weight 0.25 for the couple {p,q}$.
Assuming the claim, let's prove a lower bound of 0.25.
Let $A,B$ be the disjoint sets of weight 0.25, 0.5 respectively for the first couple and $C,D$ be the disjoint sets of weight 0.25, 0.5 resp. for the second couple.
Let $U=B \setminus \{C \cup D\}$.
One of $(B \cap C) \cup U$ and $(B \cap D) \cup U$ must have weight at least 0.25 for the first couple.
In the first case, give the books of $(B \cap C) \cup U$ to the first couple and the books of $D$ to the second couple. 
In the second case, give the books of $(B \cap D) \cup U$ to the first couple and the books of $C$ to the second couple.
This completes the proof assuming the claim.
Proof of Claim:
Let $R,S$ be the sets of weight 0.5 for $p,q$ respectively. 
Note that $\bar{R},\bar{S}$ also have weights 0.5 for $p,q$ respectively.
Since $S,\bar{S}$ is a partition of the books, we know that one of $S$ and $\bar{S}$ must have weight at least 0.5 for $p$, say $S$.
Thus $S$ has weight at least 0.5 for both $p,q$. By a similar argument, we can assume that $R$ has weight at least 0.5 for both $p,q$.
Now I show that we can take the pair $(A,B)$ be to be one of the following: $(\bar{S},S)$, $(\bar{R},R)$, $(R \cap S$, $\bar{R} \cup \bar{S})$.
For contradiction, suppose the first two pairs don't satisfy the claim.
Then $w_1(\bar{S}) \leq 0.25$ and $w_2(\bar{R}) \leq 0.25$.
This implies that $w_1(R \cap S) \geq 0.25$ and $w_2(S \cap R)\geq 0.25$, which means that the third pair satisfies the claim.