Let $A$ and $B$ be two Hermitian matrices. The famous Golden-Thompson inequality states that

$$\text{tr}(e^{A+B}) \le \text{tr}(e^Ae^B)$$

However, for determinants we have equality

$$\det(e^{A+B}) =\det(e^Ae^B)$$

I was wondering if similar results can be shown, if instead of trace and determinant, we use any of the other *fundamental scalar functions* of a matrix (e.g., trace is $\phi_1(X) :=\sum_i \lambda_i(X)$; $\phi_2(X)=\sum_{i \neq j} \lambda_i(X)\lambda_j(X)$, determinant is $\phi_n$)

PS: Please feel free to add more tags, if you deem it to be necessary.