Given three real, symmetric matrices $A\succ0$ and $B$, $C⪰ 0$.


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!



Determinant of non-symmetric sum of matrices
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.


