TITLE: Probability of symmetric difference
QUESTION [6 upvotes]: How do we prove the following inequality?
$$\mathbb{P}[A\triangle B ]\geq \max\left \{ \mathbb{P}[A-B],\mathbb{P}[B-A] \right \}$$
I already proved that
$$\mathbb{P}[A\triangle B ]=\mathbb{P}[A]+\mathbb{P}[B]-2\mathbb{P}[A\cap B]$$
$$\left | \mathbb{P}[A]-\mathbb{P}[B] \right |\leq \mathbb{P}[A\triangle B ]$$

REPLY [3 votes]: Note that
$$
A\Delta B=(A\setminus B)\cup (B\setminus A)
$$
whence
$$
(A\setminus B)\subseteq A\Delta B\quad \text{and}\quad 
(B\setminus A)\subseteq A\Delta B.
$$
In particular
$$
P(A\setminus B)\leq P(A\Delta B)
\quad
\text{and}
\quad
P(B\setminus A)\leq P(A\Delta B)
$$
so that
$$
\max\{
P(A\setminus B), P(B\setminus A)
\}
\leq P(A\Delta B)
$$