Probability of symmetric difference 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
$$\mathbb{P}[A\triangle B ]=\mathbb{P}[A]+\mathbb{P}[B]-2\mathbb{P}[A\cap B]$$
$$\left | \mathbb{P}[A]-\mathbb{P}[B] \right |\leq \mathbb{P}[A\triangle B ]$$

We have
$$P(A-B)=P(A)-P(A\cap B)$$
$$P(B-A)=P(B)-P(B\cap A)$$
Using these we can rewrite the first relation as
$$P(A\triangle B)=P(A-B)+P(B-A)$$
Since probabilities are non-negative:
$$P(A\triangle B)\ge\max(P(A-B),P(B-A))$$
