Suppose $A,A\_1,\ldots,A\_{n-2}$ (resp. $B$) are (resp. is) real positive-definite (resp. arbitrary) symmetric $n\times n$ matrices and denote by $D(\cdot,\ldots,\cdot)$ the mixed discriminant. We have the following well-known Aleksandrov-Fenchel inequality 


\begin{equation}\label{e}
D(A,B,A\_1,\ldots,A\_{n-2})^2\geq D(A,A,A\_1,\ldots,A\_{n-2})D(B,B,A\_1,\ldots,A\_{n-2}),
\end{equation}
with equality iff $A$ and $B$ are proportional.


Now my question comes: does this inequality and equality case still hold if we assume these matrices are complex Hermitian matrices rather than real symmetric ones? I guess this is the case but I am not able to find a reference. The standard textbook of Schneider only treats the real symmetric case.


Many thanks in advance!



The Aleksandrov-Fenchel inequality of mixed discriminants for Hermitian matrices
[The Van der Waerden Conjecture for Mixed Discriminants](https://arxiv.org/abs/math/0406420), by Leonid Gurvits (2004), proves a generalized Alexandrov-Fenchel inequality for semidefinite $n\times n$ Hermitian matrices: 


Theorem 5.2, with equation (21), that for the special case $\alpha=(1,1,\ldots,1)$, $\alpha^{(1)}=(2,0,1,\ldots 1)$, $\alpha^{(2)}=(0,2,1,\ldots 1)$ reduces to the desired inequality with an extra factor $\exp(-2n^n/n!)$ on the right-hand side. 


Not good enough, obviously, but I have not found a proof for the Hermitian case without this reduction factor.


