Positivity of trace of some product of matrices

Question

Consider complex square matrices $A,B$. I suspect the following inequality holds \begin{equation} \textrm{tr}\left[(A^{\dagger}A + AA^{\dagger})B^{\dagger}B - A^{\dagger}BAB^{\dagger} - BA^{\dagger}B^{\dagger}A \right] \geq 0, \end{equation} where $\dagger$ denotes hermitian conjugation. The expression inside the brackets is not positive semi-definite while the trace seems to be (I checked the statement for many copies of random complex matrices $A,B$). However, I am struggling to prove the statement. Can anyone see why the above holds or point me to a counter example?

Answer 1

This isn't true. Perhaps I was lucky and you were not. I got a counterexample on the first shot: $$ A=\pmatrix{1&0\\ 0&2},\ B=\pmatrix{1&0\\ 1&1}, $$ \begin{aligned} &(A^\dagger A + AA^\dagger )B^\dagger B - A^\dagger BAB^\dagger - BA^\dagger B^\dagger A\\ =\ &\pmatrix{2&0\\ 0&8}\pmatrix{2&1\\ 1&1}-\pmatrix{1&0\\ 2&2}\pmatrix{1&1\\ 0&2} -\pmatrix{1&0\\ 1&2}\pmatrix{1&2\\ 0&2}\\ =\ &\pmatrix{4&2\\ 8&8}-\pmatrix{1&1\\ 2&6}-\pmatrix{1&2\\ 1&6} =\pmatrix{2&-1\\ 5&-4}. \end{aligned} However, by the tracial property, we do have \begin{aligned} &\operatorname{tr}\left(A^\dagger AB^\dagger B + AA^\dagger \color{red}{BB^\dagger} - A^\dagger BAB^\dagger - BA^\dagger B^\dagger A\right)\\ =\ &\operatorname{tr}\left(AB^\dagger BA^\dagger + B^\dagger AA^\dagger B - AB^\dagger A^\dagger B - B^\dagger ABA^\dagger\right)\\ =\ &\operatorname{tr}\left[\left(AB^\dagger-B^\dagger A\right)\left(BA^\dagger-A^\dagger B\right)\right]\\ =\ &\operatorname{tr}\left[\left(BA^\dagger-A^\dagger B\right)^\dagger\left(BA^\dagger-A^\dagger B\right)\right]\ge0. \end{aligned}