Define
$$A = \frac{a_1 a_2}{a_1 - a_2}, \quad B = \frac{a_1 a_3}{a_1 - a_3}, \quad C = \frac{a_2 a_3}{a_2 - a_3}.$$
Then
$$\begin{align*} & a_1 b_1 + a_2 b_2 + a_3 b_3 \\ = & a_1 (1+A) (1+B) + a_2 (1-A) (1+C) + a_3 (1-B) (1-C) \\ = & a_1 + a_2 + a_3 + (a_1 - a_2)A + (a_1 - a_3)B \\ & + (a_2 - a_3)C + a_1 AB - a_2 AC + a_3 BC. \end{align*}$$
Some computation shows that
$$(a_1 - a_2)A + (a_1 - a_3)B + (a_2 - a_3)C = a_1 a_2 + a_1 a_3 + a_2 a_3,$$

$$\begin{aligned} & a_1AB - a_2AC + a_3BC \\ &= a_1a_2a_3 \frac{a_1^2(a_2 - a_3) + a_2^2(a_3 - a_1) + a_3^2(a_1 - a_2)}{(a_1 - a_2)(a_2 - a_3)(a_1 - a_3)} \\ &= a_1a_2a_3. \end{aligned}$$
$$Thus$$
$$\begin{aligned} & 1 + |a_1b_1 + a_2b_2 + a_3b_3| \\ &= 1 + |a_1 + a_2 + a_3 + a_1a_2 + a_1a_3 + a_2a_3 + a_1a_2a_3| \\ &\le 1 + |a_1| + |a_2| + |a_3| + |a_1a_2| + |a_1a_3| + |a_2a_3| + |a_1a_2a_3| \\ &= (1 + |a_1|)(1 + |a_2|)(1 + |a_3|). \end{aligned}$$
$$$$
The equality holds if and only if 7 real numbers $a_1$, $a_2$, $a_3$, $a_1a_2$, $a_1a_3$, $a_2a_3$, $a_1a_2a_3$ are either all non-negative or all non-positive. Note that at most one of the three numbers $a_1$, $a_2$, $a_3$ can be zero. Thus, the equality holds if and only if the three real numbers $a_1$, $a_2$, $a_3$ are all non-negative. $\square$