Consider the following matrix with nonnegative entries: $$ M=\begin{pmatrix} a & b & c & d \\ b & c & d & e \\ c & d & e & f \\ d & e & f & g \\ \end{pmatrix}. $$
Can we prove that, if each minor $2\times 2$ has nonnegative determinant, then the determinant of $M$ itself is nonnegative?
Computer experiments nowadays are cheap to run. Why didn't you generate some random matrices to see if your hypothesis is supported by numerical evidence? Anyway, here is a counterexample: $$ M = \pmatrix{ 6&4&3&3\\ 4&3&3&3\\ 3&3&3&5\\ 3&3&5&9}. $$ All $2\times2$ minors of $M$ are nonnegative, but $\det M = -14 <0$.