Suppose that $A\_i$, $i=1,\ldots,m$ are positive semidefinite Hermitian $n\times n$ matrices, with $a\_i^{(j)}$ being the $j$-th eigenvalue of $A\_i$. Let $A\_0=I$. 



> 
> **QUESTION.** Can we extend the result of M. Fiedler for $m=2$ (*Bounds for the Determinant of the Sum of Hermitian Matrices*, Proc. Amer. Math. Soc.
>  **30** (1971) 27-31) to establish the bound
>  $$
> \det \sum\_{i=0}^mA\_i \le\max\_{\sigma\_0,\dots,\sigma\_m\in S\_n}\prod\_{j=1}^n\sum\_{i=0}^m a\_i^{(\sigma\_i(j))}\,\,?
> $$
>  ($S\_n$ is the group of permutations of $n$ elements)
> 
> 
> 


**Edit**: it should be noted $A\_0$ is superfluous, since the statement would hold for singular matrices too. I didn't want the answers below to have to consider the additional case.



Determinant of arbitrary sum of positive semidefinite hermitian matrices
The previous answer was incorrect due to an incorrect use of majorization. I decided to replace it by a correct version below for completeness, and to remedy the embarrassing error (though now the answer becomes essentially equivalent to the OPs own answer, so feel free to ignore!).
$\newcommand{\da}{\downarrow} \newcommand{\ua}{\uparrow} \newcommand{\nlsum}{\sum\nolimits}$




---


Recall that for Hermitian matrices $A$ and $B$, [Lidskii's](https://terrytao.wordpress.com/tag/lidskii-inequality/) eigenvalue [majorization](https://en.wikipedia.org/wiki/Majorization) inequality implies that there exist doubly stochastic matrices $D\_1$ and $D\_2$ such that $\lambda(A+B)=D\_1\lambda(A)+D\_2\lambda(B)$.


Introduce the shorthand $a\_{m+1}=\lambda(A\_1+\cdots+A\_m)$, and $a\_j = \lambda(A\_j)$ for $1\le j\le m$. Then, applying Lidskii's result repeatedly and noting that doubly stochastic matrices are closed under multiplication, it follows that there exist doubly stochastic matrices $D\_1,\ldots,D\_m$ such that
\begin{equation\*}
 a\_{m+1} = D\_1a\_1+\cdots+D\_ma\_m.
\end{equation\*}


We wish to upper bound $\det(A\_{m+1}):=\prod\_i a\_{m+1,i}$. Consider the function
\begin{equation\*}
 f(D\_1,\ldots,D\_m) := \prod\_i (D\_1a\_1+\cdots+D\_ma\_m)\_i. % \le \left(\frac{\nlsum\_i e\_i^T(D\_1a\_1+\cdots+D\_ma\_m)}{n} \right)^n.
\end{equation\*}


We now show that $f$ is maximized over permutation matrices. First, we make the change of variables $(D\_l)\_{ij}=D\_{l,ij} = e^{u\_{l,ij}}$. Moreover, we relax the doubly-stochastic constraints on each $D\_l$ to
\begin{equation\*}
 \Omega\_n := \left\lbrace(U\_1,\ldots,U\_m) \in \mathbb{R}^{m\times n \times n}\mid \nlsum\_i e^{u\_{l,ij}} \le 1,\ \forall i,\quad \nlsum\_j e^{u\_{l,ij}} \le 1,\forall j, \quad\text{for}\ 1 \le l \le m\right\rbrace.
\end{equation\*}
Then, we consider the optimization problem
\begin{equation\*}
 \sup\_{(U\_1,\ldots,U\_m) \in \Omega\_n}\quad g(U\_1,\ldots,U\_m) := \nlsum\_{i=1}^n \log\left(\nlsum\_{lj} a\_{l,j}e^{u\_{l,ij}}\right).
\end{equation\*}
Since all the matrices $A\_1,\ldots,A\_m$ are positive definite, each $a\_{l,j}\ge0$. Thus, $g$ is a convex function of the $U\_i$. Hence its supremum will be at an extreme point of $\Omega\_n$. These points correspond to the permutation matrices. Thus, mapping back to the $D\_i$ space, we observe that $f$ will be maximized at permutation matrices. 


This reasoning immediately yields the desired inequality:
\begin{equation\*}
 \det(A\_{m+1})=\det(A\_1+\cdots+A\_m) \le \sup\_{\sigma\_1,\ldots,\sigma\_m \in S\_n}\prod\_{i=1}^n\left(\nlsum\_{l=1}^m a\_{l,\sigma\_l(i)} \right).
\end{equation\*}


