How to prove this trace matrix inequality？

Question

Given:

1. $A$ is a diagonal positive definite matrix;
2. $\operatorname{Tr}(A)=1$ and $\operatorname{Tr}(A^{2})\leq 1$;
3. $B$ is a Hermitian matrix;
4. $AB\neq BA$.

How to prove the following:

$$\operatorname{Tr}(AABB)-\operatorname{Tr}(ABAB)\leq \operatorname{Tr}(A^{2})\left[ \operatorname{Tr}(ABB)-\operatorname{Tr}(AB)\operatorname{Tr}(AB)% \right]? $$

Answer 1

This is not true. Random counterexample: $$ A=\pmatrix{0.66\\ &0.33\\ &&0.01}, \ B=\pmatrix{0&2&15\\ 2&0&10\\ 15&10&0}, $$ $\operatorname{tr}(AABB)-\operatorname{tr}(ABAB) =105.74>102.77 =\operatorname{tr}(A^2)\left[\operatorname{tr}(ABB)-\operatorname{tr}(AB)\operatorname{tr}(AB)\right]$.