A possible extension of a determinant inequality It is well known that if $A, B$ are positive semidefinite matrices, then $$\det (A+B)\ge \det A+\det B.$$ 
I am considering a possible extension of this result. Let $\mathbb{M}_m(\mathbb{M}_n)$ denote the set of $m\times m$ block matrices with each block the usual $n\times n$ matrix. 
Let  $\mathbf{A}=[A_{i,j}]_{i,j=1}^m, \mathbf{B}=[B_{i,j}]_{i,j=1}^m \in \mathbb{M}_m(\mathbb{M}_n)$ be  positive semidefinite. Define a new matrix $M=[m_{ij}]_{i,j=1}^m$ with $m_{ij}=\det (A_{i,j}+B_{i,j})-(\det A_{i,j}+\det B_{i,j})$. Is it true that $M$ is positive semidefinite?
I ran some numerical simulations, yet no counterexamples showed up (of course, numerical simulations can never be thorough). I did not find a proof even when $m=2$.
Clearly, it suffices to show $\det M\ge 0$.
 A: The claim is true. We prove it using a few block matrix manipulations. Note, in the proofs below $A \ge 0$ means $A$ is (symmetric) positive semidefinite.
$\newcommand{\trace}{\text{trace}}$

Lemma Let $X, Y \ge 0$. Then,
  \begin{equation*}
   \otimes^k (X+Y) \ge \otimes^k X + \otimes^k Y.
\end{equation*}
  Proof.
    By induction on $k$. The case $k=1$ is trivial; $k=2$ shows us the crux. Indeed,
    \begin{equation*}
    (X+Y)\otimes (X+Y) - X\otimes X - Y\otimes Y = X\otimes Y + Y \otimes X \ge 0,
  \end{equation*}
    since $X, Y \ge 0$. The general case follows similarly. 

-

Corollary. 
    Let $X, Y \ge 0$. Then, by restricting to the suitable symmetry class of tensors we get
    \begin{equation*}
    \wedge^k(X+Y) \ge \wedge^k X + \wedge^k Y
  \end{equation*}

.

Lemma.   Let $A=[A_{ij}]$ be $mn\times mn$ with $n\times n$ blocks. Suppose $A$ is symmetric, positive semidefinite. Then, for $1\le k \le n$, the $m\binom{n}{k} \times m\binom{n}{k}$ matrix $C_k := [\wedge^k A_{ij}]$ is semidefinite.

Proof.  Some reflection shows that $C_k$ is a principal submatrix of $\wedge^k A$, thus, $C_k = P_k^*(\wedge^k A)P_k \ge 0$ since $A\ge 0$ and wedge products preserve positivity.

Theorem.   Let $A=[A_{ij}] \ge 0$ and $B=[B_{ij}] \ge 0$ be $mn\times mn$ block matrices composed of $n\times n$ blocks. Define $$M_k = [\trace(\wedge^k(A_{ij}+B_{ij}))] - [\trace(\wedge^k A_{ij})] - [\trace(\wedge^k B_{ij})],$$ for any $1\le k \le n$. Then, $M_k \ge 0$.

Proof
  The Corollary above shows that $\wedge^k(A+B) \ge \wedge^k A + \wedge^k B$. Let $P_k$ be as in the second Lemma; then $$H_k = P_k^*(\wedge^k(A+B))P_k- P_k^*(\wedge^k A)P_k -P_k^*(\wedge^k B)P_k \ge 0.$$ The matrix $M_k$ is nothing but a (blockwise) partial trace of $H_k$, so that $H_k \ge 0 \implies M_k \ge 0$.

Corollary
    Let $A$ and $B$ be as above. Then,
    \begin{equation*}
    M = [\det(A_{ij}+B_{ij})] - [\det A_{ij}] - [\det B_{ij}] \ge 0.
  \end{equation*}

Proof.
  Observe that $\trace(\wedge^n X) = \det(X)$ for an $n\times n$ matrix $X$.
