trace inequality for trace(ABA*C) I have three $n \times n$ matrices $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ and I calculate the trace
\begin{gather}
 \mathrm{trace}\left(\mathbf{A} \mathbf{B} \mathbf{A}^* \mathbf{C}\right),
\end{gather}
where $\mathbf{A}^*$ denotes the conjugate transpose (Hermitian) of $\mathbf{A}$.
Matrix $\mathbf{A}$ is positive definite with $\left\|\mathbf{A}\right\|^2 \leq n$.
Matrices $\mathbf{B}$, $\mathbf{C}$ are Hermitian positive semidefinite, with $\mathbf{B} \mathbf{B}^* = \mathbf{B}$ and $\mathbf{C} \mathbf{C}^* = \mathbf{C}$, and $\left\|\mathbf{B}\right\|^2 = \left\|\mathbf{C}\right\|^2 = m < n$.
I would like to prove/disprove the following inequality:
\begin{gather}
\mathrm{trace}\left(\mathbf{A}\mathbf{B}\mathbf{A}^*\mathbf{C}\right) \leq \mathrm{trace}\left(\mathbf{A}\mathbf{A}^*\right) \mathrm{trace}\left(\mathbf{B}\mathbf{C}\right) 
\end{gather}
Any hint would be welcome.

I have found now a counterexample that shows that the considered inequality does generally not hold. Specifically, if B and C are generated as $B = Q_B Q_B^*$ and $C = Q_C Q_C^*$, where $Q_B$ and $Q_C$ are bases for orthogonal $m$-dimensional subspaces, then $\mathrm{trace}(B C)$ is zero, whereas $\mathrm{trace}(A B A^* C)$ is generally not zero.
