Let $A, B$ be Hermitian matrices. Does the following hold?
$$\det(A^{2}+B^{2}+|AB+BA|)\leq \det(A^{2}+B^{2}+|AB|+|BA|)$$
As usual, $|X|=(X^*X)^{1/2}$. Clearly, quantities on both sides are no less than $\det(A+B)^2$.
Let $A, B$ be Hermitian matrices. Does the following hold?
$$\det(A^{2}+B^{2}+|AB+BA|)\leq \det(A^{2}+B^{2}+|AB|+|BA|)$$
As usual, $|X|=(X^*X)^{1/2}$. Clearly, quantities on both sides are no less than $\det(A+B)^2$.
Consider the following matlab output:
The square root function produces a very small imaginary error in the second case so I've just stripped that away in the final calculation. Therefore, the proposed determinantal inequality does not hold.