6. Then, in: the snake $A_{1} A_{2} B_{2} B_{1}$ is small, and it is known that $A_{1} A_{2} = 2a, A_{1} B_{2} < \sqrt{2} a$. Taking $A_{1} A_{2}$ as the major axis, construct an ellipse $C$ such that the focal distance of $C$ is $\mathrm{f} = \sqrt{2} A_{1} B_{1}$. On $C$, take a point $P$ (not coinciding with the endpoints). Let the intersections of the lines $P B_{1}$ and $P B_{2}$ with $A_{1} A_{2}$ be $M_{1}$ and $M_{2}$, respectively. Then $A_{1} M_{2}^{2} + A_{2} M_{1}^{2} =$
6. $4 a^{2}$.

Let the line through $A_{1}, A_{2}$ be the $x$-axis, and the perpendicular bisector of $A_{1} A_{2}$ be the $y$-axis to establish a rectangular coordinate system. Suppose $A_{1} B_{1}=2 b$, then the equation of the ellipse $C$ is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{2 b^{2}}=1$.

Let $P(a \cos \theta, \sqrt{2} b \sin \theta), \theta \in(0,2 \pi), \theta \neq \frac{\pi}{2}, \pi, \frac{3 \pi}{2}$, and $B_{1}(-a, 2 b), B_{2}(a, 2 b)$, then the equations of $P B_{1}$ and $P B_{2}$ are respectively
$$
\begin{array}{l}
P B_{1}: \frac{y-2 b}{\sqrt{2} b \sin \theta-2 b}=\frac{x+a}{a \cos \theta+a}, \\
P B_{2}: \frac{y-2 b}{\sqrt{2} b \sin 2 \theta-2 b}=\frac{x-a}{a \cos \theta-a} . \\
\text { Let } y=0, \text { then } \\
x_{M_{1}}=\frac{2 a(\cos \theta+1)}{2-\sqrt{2} \sin \theta}-a, \\
x_{M_{2}}=\frac{2 a(\cos \theta-1)}{2-\sqrt{2} \sin \theta}+a . \\
\therefore\left|A_{1} M_{2}\right|^{2}+\left|A_{2} M_{1}\right|^{2}=\left(x_{M_{2}}+a\right)^{2}+\left(x_{M_{1}}-a\right)^{2} \\
=\left[\frac{2 a(\cos \theta-1)}{2-\sqrt{2} \sin \theta}+2 a\right]^{2}+\left[\frac{2 a(\cos \theta+1)}{2-\sqrt{2} \sin \theta}-2 a\right]^{2} \\
=\cdots=4 a^{2} .
\end{array}
$$