Given three real, symmetric matrices $A\succ0$ and $B$, $CâȘ° 0$.
How can it be shown that:
$$\det(A^2+AB+AC) \leq \det(A^2 +BA +AC+BC) ? \qquad (\star)$$
Where $A^2$ is symmetric and positive definite. Eigenvalues of $BA$, $AC$, and $BC$ are all $> 0$, but symmetry is lost.
Thank you!
Best Answer
Because the original question has changed so much, I am writing a new answer.
The key point to recognize is that you are trying to prove a submodularity property. Indeed, we see that we may equivalently prove \begin{equation*} \log\det(A) + \log\det(A+B+C) \le \log\det(A+B) +\log\det(A+C). \end{equation*} One common way to verify submodularity is to prove the diminishing marginals property: in our case, it amounts to showing that for a fixed $B \succeq 0$, the function \begin{equation*} f(A) := \log\det(A+B)-\log\det(A) \end{equation*} is monotonically decreasing (since we are dealing with hermitian positive definite matrices, this means $f(A) \le f(C)$ if $C \succeq A$ in the semidefinite order).
To verify this, simply check if $\nabla f(A) \preceq 0$ for all $A$. But this is easy since
\begin{equation*} \nabla f(A) = (A+B)^{-1} - A^{-1} \preceq 0, \end{equation*} where the latter inequality follows as the map $X \mapsto X^{-1}$ is well-known to be (operator) decreasing.