Let $M_{1},M_{2},M_{3},M_{4} \in M_{3}(\Bbb{R})$. If $M_{1} = \begin{bmatrix} a_{1} + 1 & a_{3} & b_{2}\\ a_{2} & a_{1}+1 & b_{3}\\ a_{3} & a_{2} & b_{1}+1\end{bmatrix}, M_{2} = \begin{bmatrix} a_{1} + 1 & b_{3} & b_{2}\\ a_{2} & b_{1}+1 & b_{3}\\ a_{3} & b_{2} & b_{1}+1\end{bmatrix}$
$M_{3} = \begin{bmatrix} a_{1} & 1 & b_{2}\\ a_{2} & 1 & b_{3}\\ a_{3} & 1 & b_{1}\end{bmatrix}, M_{4} = \begin{bmatrix} b_{1} & 1 & a_{2}\\ b_{2} & 1 & a_{3}\\ b_{3} & 1 & a_{1}\end{bmatrix}$
$\textbf{Assumption:}$ Suppose the absolute values of all the eigenvalues of $M_{1}$ and $M_{2}$ are less than one.
Then to prove that either $\det(M_{3}) >0$ or $\det(M_{4})>0$.
Thoughts:
a) Let $\tau_{M},\delta_{M}$ denote the trace and eigenvalue of matrix $M$. The absolute value of all the eigenvalues of $M$ are less than one if and only if $1+\tau_{M}+\delta_{M}>0, 1-\tau_{M}+\delta_{M}>0,$ and $ \delta_{M}<1$. I now think this statement can work only for $2 \times 2$ matrix.
b) Maybe geometrically we can think of the determinant as a triple product. $\det(\textbf{a},\textbf{b},\textbf{c}) = \textbf{a}\,.(\textbf{b} \times \textbf{c})$.