Problem 1234

Prove: $1+\left|a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\right| \leqslant\left(1+\left|a_{1}\right|\right)\left(1+\left|a_{2}\right|\right)\left(1+\left|a_{3}\right|\right)$, and identify the conditions under which equality holds.
Proof Let $A=\frac{a_{1} a_{2}}{a_{1}-a_{2}}, B=\frac{a_{1} a_{3}}{a_{1}-a_{3}}, C=\frac{a_{2} a_{3}}{a_{2}-a_{3}}$, then
$$\begin{aligned} & a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}=a_{1}(1+A)(1+B)+a_{2}(1-A)(1+C)+a_{3}(1-B)(1-C) \\ = & a_{1}+a_{2}+a_{3}+\left(a_{1}-a_{2}\right) A+\left(a_{1}-a_{3}\right) B+\left(a_{2}-a_{3}\right) C+a_{1} A B-a_{2} A C+a_{3} B C \end{aligned}$$

By calculation, we get $\left(a_{1}-a_{2}\right) A+\left(a_{1}-a_{3}\right) B+\left(a_{2}-a_{3}\right) C=a_{1} a_{2}+a_{1} a_{3}+a_{2} a_{3}$,
$$a_{1} A B-a_{2} A C+a_{3} B C=a_{1} a_{2} a_{3} \frac{a_{1}^{2}\left(a_{2}-a_{3}\right)+a_{2}^{2}\left(a_{3}-a_{1}\right)+a_{3}^{2}\left(a_{1}-a_{2}\right)}{\left(a_{1}-a_{2}\right)\left(a_{2}-a_{3}\right)\left(a_{1}-a_{3}\right)}=a_{1} a_{2} a_{3} .$$

Therefore, $1+\left|a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\right|=1+\left|a_{1}+a_{2}+a_{3}+a_{1} a_{2}+a_{1} a_{3}+a_{2} a_{3}+a_{1} a_{2} a_{3}\right|$
$$\begin{array}{l} \leqslant 1+\left|a_{1}\right|+\left|a_{2}\right|+\left|a_{3}\right|+\left|a_{1} a_{2}\right|+\left|a_{1} a_{3}\right|+\left|a_{2} a_{3}\right|+\left|a_{1} a_{2} a_{3}\right| \\ =\left(1+\left|a_{1}\right|\right)\left(1+\left|a_{2}\right|\right)\left(1+\left|a_{3}\right|\right) . \end{array}$$

Equality holds if and only if the seven real numbers $a_{1}, a_{2}, a_{3}, a_{1} b_{1}, a_{2} b_{2}, a_{3} b_{3}, a_{1} a_{2} a_{3}$ are all non-negative or all non-positive. Noting that at most one of $a_{1}, a_{2}, a_{3}$ can be 0, thus, equality holds if and only if $a_{1}, a_{2}, a_{3}$ are all non-negative.