Notice that $0 < a, b \le 1$ and by the substitution $m = A - 3a$ and $n = B - 3b$, we have $|m| \le 1 - a$ and $|n| \le 1 - b$. It suffices to show that $$ \left|\frac{mn}{3} + mb + na \right | \le 1 - ab $$Using the inequality $|x+y| \le |x| + |y|$ for $x,y \in \mathbb{C}$, we have $$ \left|\frac{mn}{3} + mb + na \right | \le \left|\frac{mn}{3}\right| + |m|b + |n|a \le \frac{(1-a)(1-b)}{3} + a(1-b) + b(1-a) \le 1 - ab $$We complete our proof here.