Question 75
Suppose the real numbers $a$, $A$, $b$, $B$ satisfy the inequalities:
$$
|A - 3a| \le 1 - a, \quad |B - 3b| \le 1 - b,
$$
and $a$, $b$ are positive. Prove that $\left| \frac{AB}{3} - 3ab \right| \le 1 - ab$.
$$
|A - 3a| \le 1 - a, \quad |B - 3b| \le 1 - b,
$$
and $a$, $b$ are positive. Prove that $\left| \frac{AB}{3} - 3ab \right| \le 1 - ab$.