Matrix inequality : trace of exponential of Hermitian matrix

Question

I want to know whether the following inequality holds or not. \begin{align} (\mathrm{Tr}\exp[(A+B)/2])^2\leq(\mathrm{Tr}\exp A)(\mathrm{Tr}\exp B)\tag{1} \end{align} where $A, B$ are Hermitian matrices of the dimension $D$.

Note that if $A$ and $B$ commute, we can see (1) holds using the simultaneously diagonalizing basis and Cauchy-Schwarz inequality. The problem is the case where $A$ and $B$ do not commute.

Answer 1

You can prove it using the Golden-Thompson inequality $Tr (e^{A+B}) \leqslant Tr(e^{A} e^{B})$ and then applying the Cauchy-Schwarz inequality.