Determinant of non-symmetric sum of matrices 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!
 A: 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.
A: None of the conjectured inequalities hold.
This answer contains three counterexamples. The first one is to $(\star)$, while the second and third ones (below the line) refer to previous inequalities conjectured by the OP.
\begin{equation*}
  A=\begin{bmatrix} 5 & 5\\\\ 5 & 5\end{bmatrix},\quad
  B=\begin{bmatrix} 8 &4 \\\\ 4 & 2\end{bmatrix},\quad
  C=\begin{bmatrix} 8 &6 \\\\ 6 & 5\end{bmatrix}.
\end{equation*}
Then $A^2+AB+AC$ is a rank-1 matrix, so its determinant is zero, while $A^2+AB+AC+BC=\begin{bmatrix}268 & 203\\\\ 224 & 169\end{bmatrix}$, so its determinant is $-180$.  
The structure of this counterexample and of the other ones below is to setup matrices so that the left hand side becomes a rank-1 matrix, which will have determinant zero. Then, one can adjust the other terms to violate inequalities in any direction.
EDIT I'm adding explicit matrices where $A,B,C \succ 0$, but still we have a counterexample to quell the OP's insistence ;-)
\begin{equation*}
A=\begin{bmatrix}11&12&7\\\\ 12 & 14 & 8\\\\ 7 & 8 & 11\end{bmatrix},\ 
B=\begin{bmatrix}19 &14&7\\\\ 14& 14&6\\\\ 7&6&3\end{bmatrix},\
C=\begin{bmatrix}17&17&16\\\\ 17&19&17\\\\ 16&17&17\end{bmatrix}
\end{equation*}
For these matrices, $\det(A^2+AB+AC) \approx 2.35\times 10^5$, while $\det(A^2+AB+AC+BC) \approx 4.67 \times 10^4$.

A particularly cute counterexample for your last question (edit: where $\det(A+B) \ge \det(A)+\det(B)$ holds for non symmetric matrices with positive eigenvalues) is the following:
\begin{equation*}
 A = \begin{bmatrix}
     0.5 &         0 &         0&         0\\\\
    1    &    0.5 &        0 &        0\\\\
    1    &    1 &    0.5 &        0\\\\
    1    &    1 &   1   & 0.5
  \end{bmatrix},\quad B = A^T.
\end{equation*}
Then, $\det(A+B)=0$, but $\det(A)+\det(B) = 1/8$.
The updated question, whether $\det(XY+YZ) \ge \det(XY)+\det(YZ)$ holds is also false. Here is a nice counterexample.
\begin{eqnarray*}
  X = \begin{bmatrix}
   2 & 2\\\\
   2 & 2
  \end{bmatrix},\quad Y = \begin{bmatrix}
   5 & 5\\\\
   5 & 10
   \end{bmatrix},\quad Z = \begin{bmatrix}
   10 & 4\\\\
   4  & 2
   \end{bmatrix}.
\end{eqnarray*}
With this choice, $\det(XY+YZ) = -300$, while $\det(XY)+\det(YZ)=100$.
