Show determinant is non-negative Let $A,B \in M_{2}(\mathbb R)$ . Show that $\det((AB+BA)^4 + (AB-BA)^4)\geq 0$
My attempt: expression becomes $\det(2(M-N)^2+16MN)$ where $M=(AB)^2$ and $N=(BA)^2$.
Not sure how to continue from here.
Any hints appreciated.

I believe that approaching the problem is easier by showing the Positive Semi-Definiteness of M and N. This way you can generalize it to $A,B \in R^{n \times n}$.
Expanding the square of matrix $AB$ and $BA$ according to this source, $M = (AB)^2 = ABB^TA^T, N = (BA)^2 = BAA^TB^T$ are symmetric. As the result their eigenvalues are real.
Moreover, M and N are PSD (shown at the end of the answer). The determinant of a PSD matrix is greater or equal to zero. Thus, it is enough to show that $2(M - N)^2 + 16MN = 2M^2 + 12MN + 2N^2$ is PSD. And this is trivial since it is a summation and product of symmetric PSD matrices.
M is PSD if $x^TMx \geq 0$ $\forall x \in R^n$ $\implies x^TABB^TA^Tx = \sum_{i=1}^n\langle x,(AB)_i\rangle^2 \geq 0$ where $(AB)_i$ is $i$th column of matrix $AB$.
