Prove that among $b_{1}^{2}-a_{1}^{2}, b_{2}^{2}-a_{2}^{2}, b_{3}^{2}-a_{3}^{2}$, there must be two that are both not greater than zero or both not less than zero, i.e., $\left(b_{2}^{2}-a_{2}^{2}\right)\left(b_{3}^{2}-a_{3}^{2}\right) \geqslant 0,\left(b_{1}^{2}-a_{1}^{2}\right)\left(b_{3}^{2}-a_{3}^{2}\right) \geqslant 0,\left(b_{1}^{2}-a_{1}^{2}\right)\left(b_{2}^{2}-a_{2}^{2}\right) \geqslant 0$ must have one that holds.

Without loss of generality, assume $\left(b_{2}^{2}-a_{2}^{2}\right)\left(b_{3}^{2}-a_{3}^{2}\right) \geqslant 0$, i.e., $b_{2}^{2} b_{3}^{2} \geqslant a_{3}^{2} b_{2}^{2}+a_{2}^{2} b_{3}^{2}-a_{2}^{2} a_{3}^{2}$, then
$$\begin{aligned} \left(b_{2}^{2}+a_{1}^{2}+a_{3}^{2}\right)\left(b_{3}^{2}+a_{1}^{2}+a_{2}^{2}\right) \geqslant & \left(a_{3}^{2} b_{2}^{2}+a_{2}^{2} b_{3}^{2}-a_{2}^{2} a_{3}^{2}\right)+\left(a_{1}^{2}+a_{3}^{2}\right) b_{3}^{2}+ \\ & \left(a_{1}^{2}+a_{2}^{2}\right) b_{2}^{2}+\left(a_{1}^{2}+a_{3}^{2}\right)\left(a_{1}^{2}+a_{2}^{2}\right)= \\ & \left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(a_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right) \end{aligned}$$

Therefore,
$$\begin{array}{l} \left(b_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(a_{1}^{2}+b_{2}^{2}+a_{3}^{2}\right)\left(a_{1}^{2}+a_{2}^{2}+b_{3}^{2}\right) \geqslant \\ \left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(b_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(a_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right) \geqslant \\ \left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\right)^{2} \end{array}$$

From the above proof, it is easy to see that equality in (43) holds if and only if $a_{1}=b_{1}, a_{2}=b_{2}, a_{3}=b_{3}$.