Showing that $|A\cap B|/|A \cup B| + |B\cap C|/|B \cup C| - |A\cap C|/|A \cup C| \leq 1$ for finite sets $A,B,C$.

If $A$, $B$ and $C$ are finite sets, prove that $$ \frac{|A\cap B|}{|A \cup B|} + \frac{|B\cap C|}{|B \cup C|} - \frac{|A\cap C|}{|A \cup C|} \leq 1. $$

It seem's simple, but I tried it for a long time and cannot get it out. Maybe I can use some optimization methods to calculate it, but that's not what I want...

I'm referring to the Venn diagram above. The indicated variables denote the numbers of elements in the particular component, and are integers $\geq0$.

We have to prove that $$\eqalign{&{z+y_3\over x_1+x_2+y_1+y_2+y_3+z}+{z+y_1\over x_2+x_3+y_1+y_2+y_3+z}\cr &\qquad\leq1+{z+y_2\over x_1+x_3+y_1+y_2+y_3+z}\ .\cr}\tag{1}$$ During the proof we shall use several times the fact that the function $$t\mapsto{a+t\over b+t}\qquad(t>0)$$ is monotonically increasing when $0\leq a\leq b$.

As $x_2$ is not appearing on the RHS of $(1)$ we may assume $x_2=0$. Furthermore, the LHS of $(1)$ decreases when $y_2$ increases, whereas the mentioned principle shows that the RHS of $(1)$ is an increasing function of $y_2$. It follows that we may assume $y_2=0$ as well.

Put $y_1+y_3+z=:s$. Then we have to prove $${s-y_1\over s+x_1}+{s-y_3\over s+x_3}\leq 1+{s-y_1-y_3\over s+x_1+x_3}\ .\tag{2}$$ Here the RHS can be rewitten as $${s-y_1+x_3\over s+x_1+x_3}+{s-y_3+x_1\over s+x_3+x_1}\ .$$ Now by the aforementioned principle we have $${s-y_1\over s+x_1}\leq{s-y_1+x_3\over s+x_1+x_3}\ ,$$ and similarly for the second parts. This proves $(2)$, which is the "worst case" of $(1)$.