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
$$\eqalign{&amp;{z+y_3\over x_1+x_2+y_1+y_2+y_3+z}+{z+y_1\over x_2+x_3+y_1+y_2+y_3+z}\cr
&amp;\qquad\leq1+{z+y_2\over x_1+x_3+y_1+y_2+y_3+z}\ .\cr}\tag{1}$$
During the proof we shall use several times the fact that the function
$$t\mapsto{a+t\over b+t}\qquad(t&gt;0)$$
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 $(1)$.
Let $|A|= a,\quad |B|= b,\quad |C|= c \quad$ then

$0 \le |A \cap B|=n_1 \le a$ or $b$ $\quad $ And $\quad $ $a$ or $b \le |A \cup B|=u_1 \le a+b $

$\implies n_1 \le u_1$

$0 \le |B \cap C|=n_2 \le b$ or $c$ $\quad $ And $\quad $ $b$ or $c \le |B \cup C|=u_2 \le b+c$

$\implies n_2 \le u_2$

$0 \le |A \cap C|=n_3 \le a$ or $c$ $\quad $ And $\quad $ $a$ or $c \le |A \cup C|=u_3 \le a+c$

$\implies n_3 \le u_3$

So we have 

$\frac{n_1}{u_1} + \frac{n_2}{u_2} - \frac{n_3}{u_3} = \frac{n_1u_2u_3 + u_1n_2u_3 - u_1u_2n_3}{u_1u_2u_3}$

Case-1:  If $n_i = u_i \quad \forall 1 \le i \le 3$, then

$\frac{u_1u_2u_3 + u_1u_2u_3 - u_1u_2u_3}{u_1u_2u_3} = 1$

Case-2: If $n_i &lt; u_i \quad \forall 1 \le i \le 3$, then

$\frac{n_1u_2u_3 + u_1n_2u_3 - u_1u_2n_3}{u_1u_2u_3} &lt; \frac{u_1u_2u_3 + u_1u_2u_3 - u_1u_2u_3}{u_1u_2u_3}$

$\frac{n_1u_2u_3 + u_1n_2u_3 - u_1u_2n_3}{u_1u_2u_3} &lt; 1$