This is a problem I encountered in my studies of matrix theory:
We are given two symmetric real matrices, $ A,B $ , both of order $ n \times n $, and we are asked to show the following:
If all minors of $ A, B $ are non negative then so are all minors of the product $ AB $.
If all minors of $ A, B $ are positive then so are all minors of the product $ AB $.
Of course, this theorem stated more succinctly is "the product of totally-non negative matrices is totally-non negative". Now, I really do not know how to tackle this one, I thought maybe some determinant identity or matrix identity I am unfamiliar with is the clue. I thank all helpers.
Both claims follow from the fact that each minor of $AB$ equals a sum of products of the form $\left(\text{a minor of }A\right) \cdot \left(\text{a minor of }B\right)$ (where, unlike the original poster, we allow $0 \times 0$-minors; note that a $0 \times 0$-minor is always equal to $1$ and therefore positive). This fact, in turn, follows from Corollary 7.182 in my Notes on the combinatorial fundamentals of algebra (which is really just the Cauchy-Binet formula, with a superficial restriction on rows and columns). (The numbering might change in the future. If it does, look for Corollary 7.482 in the frozen version of 10 January 2019.)
The symmetry of the matrices $A$ and $B$ is a useless requirement.