When does a strictly diagonally dominant matrix have dominant principal minors? $A$ is an $N\times N$ matrix with diagonal elements $a_{ii}=1-s_{i}$, and off diagonal elements $a_{ij}=s_{i}w_{ij}$ for $i≠j$. Assume $0≤s_{i}<1/2$ and $\sum_{j≠i}w_{ij}=1$ for all $i$ and $0≤w_{ij}≤1$. As $1-s_{i}>∑_{j≠i}a_{ij}=s_{i}∑_{j≠i}w_{ij}$ for all $i$, matrix $A$ is strictly row diagonally dominant. Thus, $\det(A)>0$ and has positive principal minors, $M_{ii}$ where $ii$ is the submatrix from eliminating row $i$ and column $i$. (see Tsatsomeros 2002). There exists an inverse matrix $B=A^{-1}$. 
I want to show that the diagonal elements $b_{ii}=\frac{M_{ii}}{\det(A)} > 1$. I am able to show this for $N=3$ and $4$. For example, by expanding across the top gives $b_{11}>1$ as $M_{11}>-ω_{12}M_{12}+ω_{13}M_{13}-ω_{14}M_{14}$ , which follows from $M_{11}>‖M_{1j}‖_{j≠1}$. Here the results follows if the principal minor is dominant in the sense of being larger than the minors along the same row.
I do not know how to extend this to $N>4$ and have not found the result in my imperfect search of the literature. 

There is a simple proof, based on Fiedler's inequality, if your matrix is symmetric. 
If A is symmetric then A is positive definite. By Fiedler's inequality $A\circ A^{-1}-Id$ is positive semidefinite, where $A\circ A^{-1}$ stands for the Hadamard product of $A$ by $A^{-1}$.
Since $A_{ii}=1-s_i<1$ and $A_{ii}(A^{-1})_{ii}-1\geq 0$, because $A\circ A^{-1}-Id$ is positive semidefinite, then $(A^{-1})_{ii}>1$.
