For three symmetric positive semidefinite matrices $A, B,C$, I am trying to figure out if the following inequality holds, at least in some cases:


$$ \operatorname{tr} \left( A e^{B+C} \right) \leq \operatorname{tr} \left( A e^B e^C \right) $$


Note that if $A=I$ then this is the [Golden-Thompson inequality](https://en.wikipedia.org/wiki/Golden%E2%80%93Thompson_inequality):


$$\operatorname{tr} \left( e^{B+C} \right) \leq \operatorname{tr} \left( e^B e^C \right)$$


I wonder if the first inequality is true or if adding certain assumptions (such as $A≼I$ i.e. $I−A$ is positive semidefinite) will suffice to prove it. I have tried searching for similar inequalities with no luck.



An extension of the Golden-Thompson inequality
The difficulties with generalizations of the Golden-Thompson inequality to three matrices arise because the trace of a product of three positive symmetric matrices is in general not positive; unlike the trace of the product of two positive symmetric matrices, which is positive: ${\rm tr}\,e^B e^C={\rm tr}\,XX^t=\sum\_{n,m}X\_{nm}^2>0$, with $X=e^{B/2}e^{C/2}$ and $X^t$ the transpose of $X$.


So while ${\rm tr}\,Ae^{B+C}$ is a positive number, ${\rm tr}\,Ae^Be^C$ can be negative.


There do exist three-matrix generalizations of the Golden-Thompson inequality, but they take an entirely different form, see [A survey of certain trace inequalities](http://www.renyi.hu/%7Epetz/pdf/64.pdf), equation 21.


