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.



Matrix inequality : trace of exponential of Hermitian matrix
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.


