IRL ABooklet

Observe that $a$, $b \in (0, 1]$. Let $u = A - 3a$, $v = B - 3b$. Then $|u| \le 1 - a$, $|v| \le 1 - b$, and $A = u + 3a$, $B = v + 3b$, so that $AB = uv + 3(ub + va) + 9ab$. Hence
$$\begin{align*} |AB - 9ab| &= |uv + 3(ub + va)| \\ &\le |u||v| + 3(b|u| + a|v|) \\ &\le (1-a)(1-b) + 3(a(1-b) + b(1-a)) \\ &= 1 + 2(a+b) - 5ab \\ &= 3(1-ab) - 2(1-a-b+ab) \\ &\le 3(1-ab) \end{align*}$$
since
$1 - a - b + ab = (1 - a)(1 - b) \geq 0.$
Whence, dividing by 3, it follows that
$$\left| \frac{AB}{3} - 3ab \right| \leq 1 - ab.$$

When we add the two equations
$$(1-a)(1+b) = 1-ab-a+b \quad \text{and} \\ (1+a)(1-b) = 1-ab+a-b \quad \text{we obtain}$$
$$(1 - a)(1 + b) + (1 + a)(1 - b) = 2(1 - ab). \qquad (9)$$
Similarly, adding
$$(A - 3a)(B + 3b) = AB - 9ab - 3aB + 3Ab \quad \text{to} \\ (A + 3a)(B - 3b) = AB - 9ab + 3aB - 3Ab \quad \text{gives}$$
$$(A - 3a)(B + 3b) + (A + 3a)(B - 3b) = 2(AB - 9ab). \qquad (10)$$
Since $0 \le |A - 3a| \le 1 - a$, we have $0 < a \le 1$ and $A - 3a \le |A - 3a| \le 1 - a$ and so $A \le 1 + 2a \le 3$. But also $3a - A \le |A - 3a| \le 1 - a$, which gives $-1 < 4a - 1 \le A$, hence $|A| \le 3$. This implies $|A + 3a| \le |A| + 3a \le 3(1+a)$. In a similar way we obtain $|B + 3b| \le 3(1+b)$. Using (10), the triangle inequality and the assumption, we now see that
$$2|AB - 9ab| \le |A - 3a| \cdot |B + 3b| + |A + 3a| \cdot |B - 3b| \\ \le 3(1 - a)(1 + b) + 3(1 + a)(1 - b) = 6(1 - ab),$$
where we have used (9) at the end. Division by 6 yields the desired inequality.