Let $A$ and $B$ be two Hermitian matrices. The famous [Golden-Thompson inequality](http://en.wikipedia.org/wiki/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.



Extensions to the Golden-Thompson inequality?
This is theorem IX.3.5 in "Matrix analysis" by R. Bhatia (Graduate Texts in Mathematics, 169). See also corollary IX.3.6 and theorem IX.3.7. The Golden-Thompson inequality holds when $Tr$ is replaced with a function $f$ which satisfies $f(XY)=f(YX)$ and $|f(X^{2m})|\le f(|XX^{\ast}|^m)$ for all $m\geq 1$. Such functions can be the elementary symmetric functions in the eigenvalues as in your question, the product of the $k$ largest eigenvalues (in absolute value) etc.


