Triangular inequality and metrics 
Prove that the metrics $\rho(A,B)$ defined by :
$$
\rho(A,B) = 
\begin{cases}
\dfrac{\mathbb P(A \bigtriangleup B)}{\mathbb P(A \bigcup B)}  & \text{if} 
 \space \mathbb P(A \bigcup B) \neq 0 \\
0  & \text{if} \space \mathbb P(A \bigcup B) = 0 \\
\end{cases}
$$
satisfy the triangular inequality.

What i was thinking was to use the fact that $ x \rightarrow \frac{a+x}{b+x}$ when a < b and $x \geq -a$ is increasing to swith from any C to subevent of $A \bigcup B$ and then normalizing the probability function, assume that $\mathbb P(A \bigcup B) = 1 $ but i don't know how to proceed.
 A: Define
$$
\newcommand{\c}[1]{\overline{#1}\,}
\begin{align}
p_1 = P(A\,\c B\c C) && p_2 = P(\c A B\,\c C) && p_4 = P(\c A \c B C),\\
p_3 = P(A B \c C) && p_5 = P(A\c B C) && p_6 = P(\c A B C),\\
&& p_7 = P(ABC)
\end{align}
$$
I came up with this notation by representing the regions of the Venn diagram for $A,B,C$ as binary sequences of length $3$, then interpreting these binary sequences as positive integers in binary.
Then
$$
\begin{align}
\rho(A,B) 
  &=\frac{p_1+p_2+p_5+p_6}{p_1+p_2+p_3+p_5+p_6+p_7}
\\\\&\le \frac{(p_1+p_2+p_5+p_6)+2p_3+2p_4}{(p_1+p_2+p_3+p_5+p_6+p_7)+p_4}
\\\\&= \frac{p_1+p_3+p_4+p_6}{p_1+p_2+p_3+p_4+p_5+p_6+p_7}+\frac{p_2+p_3+p_4+p_5}{p_1+p_2+p_3+p_4+p_5+p_6+p_7}
\\\\&\le \frac{p_1+p_3+p_4+p_6}{p_1+p_3+p_4+p_5+p_6+p_7}\hspace{1cm}+\frac{p_2+p_3+p_4+p_5}{p_2+p_3+p_4+p_5+p_6+p_7}
\\\\&=\rho(A,C)+\rho(B,C)
\end{align}
$$
The first inequality is the only step requiring explanation. I am using the fact that if $x,y,z,w$ are positive numbers, then
$$
\frac xy < \frac zw \implies \frac xy < \frac{x+z}{y+w} < \frac zw
$$
In this case, $x=(p_1+p_2+p_5+p_6)$, $y=(p_1+p_2+p_3+p_5+p_6+p_7)$, $z=2p_3+2p_4$, and $w=p_4$.
