Probability of symmetric difference

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: $$P(A\triangle B)\ge\max(P(A-B),P(B-A))$$

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 $P(B\setminus A)$.

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 $$ \max\{ P(A\setminus B), P(B\setminus A) \} \leq P(A\Delta B) $$