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

Hint: Prove that $$P(A\Delta B)=P(A\setminus B)+P(B\setminus A)$$
Therefore, $P(A\Delta B)$ is no less than $P(A\setminus B)$ and no less than $P(B\setminus A)$.
