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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?

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  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila Sep 11 at 17:52
  • $\begingroup$ "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:-) $\endgroup$ – TonyK Sep 11 at 18:10
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As Asaf Karagila already noted, this is not a set theory question (nor a combinatorial one, I'd say), since it basically follows from an algebraic approach.

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.

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  • $\begingroup$ This is beautiful! It never occurred to me to consider those extra cardinals. You have ended my frustration, sir. $\endgroup$ – Castor Sep 11 at 18:38

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