Let $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$, $\mathbf{D}$ be four (generally non-commuting) positive semidefinite matrices of same size.
I want to show that (or find a counterexample to) $$ \operatorname{tr}\Bigl\{\mathbf{A}\mathbf{B}\mathbf{C}\mathbf{B}\mathbf{A}\mathbf{D}-\mathbf{A}\mathbf{B}\mathbf{C}\mathbf{D}-\mathbf{A}\mathbf{D}\mathbf{C}\mathbf{B}+\mathbf{C}\mathbf{D}\Bigr\} \geq 0. $$
If the matrices all commute, then the property is easy to show, because we can factorize $\mathbf{C}\mathbf{D}$ and obtain $$ \operatorname{tr}\Bigl\{\mathbf{C}\mathbf{D}\bigl(\mathbf{A}^2\mathbf{B}^2-2\mathbf{A}\mathbf{B}+\mathbf{I}\bigr)\Bigr\} = \operatorname{tr}\Bigl\{\mathbf{C}\mathbf{D}\bigl(\mathbf{A}\mathbf{B}-\mathbf{I}\bigr)^2\Bigr\} \geq 0 $$
Note: I'm only interested in cases where $\lambda_{\text{max}}(\mathbf{A}\mathbf{B}) \leq 1$ where $\lambda_{\text{max}}(\cdot)$ denotes the largest eigenvalue. However, I don't think this assumption is even needed.
Using the cyclic property of trace, you may rewrite the trace as $\operatorname{tr}(BADABC - DABC - BADC + DC)$ and in turn $\operatorname{tr}\left[(BA-I)D(AB-I)C\right]$. Since $C$ and $D$ are positive semidefinite, the resulting trace is further equal to $\operatorname{tr}\left[C^{1/2}(AB-I)^\ast D(AB-I)C^{1/2}\right]$, which is nonnegative. Note that as long as $A,B$ are Hermitian, whether they are PSD is unimportant.