# Curious question about cardinality of finite sets

Suppose we are given three finite, non-empty sets $$A, B, C$$ such that $$|C|\leq |A| \leq |B|$$. Is it always true that $$\frac{|A\cap C|}{|A|} + \frac{|B\cap C|}{|B|} \leq 1 + \frac{|A\cap B|}{|B|}$$?

I have tried countless examples for days and I haven't found any counterexample so far... This is just a curious question. But if it's true, there has to be a proof, right?

• Since these cardinals are necessarily finite (otherwise the division has no meaning), the question is not a set theory question, but rather a combinatorial question. – Asaf Karagila Sep 11 at 17:52
• "But if it's true, there has to be a proof, right?" Not necessarily. Gödel's theorems guarantee the existence of statements that are true but not provable. But in this simple case, I think it is safe to say that yes, if it's true then there has to be a proof. I'm working on it:-) – TonyK Sep 11 at 18:10

Define $$|A|=a, \ |B|=b, \ |C|=c$$ and $$|A\cap B\cap C|=x.$$ Also, let $$|(A\cap B)\setminus C|=r,$$ $$|(A\cap C)\setminus B|=s$$ and $$|(B\cap C)\setminus A|=t.$$ We can now rewrite the given inequality as $$\frac{s+x}{a}+\frac{t+x}{b}\leq 1+\frac{r+x}{b}\implies \frac{s+x}{a}+\frac{t-r}{b}\leq 1\implies b(s+x)+a(t-r)\leq ab.$$ Since all the variables are non-negative and $$c\leq a\leq b,$$ we have that $$b(s+x)+a(t-r)\leq b(s+x)+at\leq b(s+x)+bt=b(s+t+x).$$ However, notice that $$s+t+x\leq c$$ since $$(A\cap C)\setminus B, \ (B\cap C)\setminus A$$ and $$A\cap B\cap C$$ are disjoint subsets of $$C,$$ thus $$b(s+t+x)\leq bc\leq ab,$$ as desired.