An inequation about real pairwise commuting matrices

Question

Let A and B be two real $n\times n$ matrices such that $AB=BA$. It’s known that $\det(A^2+B^2)\geq 0$.
I wonder if it’s true that:
For $k$ pairwise commuting real matrices $A_1,\cdots,A_k$,we have: $\det(\sum_{i=1}^{k}A_{i}^2)\geq0$.

Answer 1

We can generalize the required result as follows

Let $(A_i)_{1≤i≤p}$ be $n$-by-$n$ real matrices that are simultaneously triangularizable over $\mathbb{C}$. Show that $\det(\sum_{i=1}^p {A_i}^2)≥0$.

I raised this problem on Image, the bulletin of the ILAS; cf. (statement and solution) p.37, pb. 56.2 in

https://www.ilasic.org/IMAGE/IMAGES/image58.pdf

A reader provided a solution (cf. the above reference) that uses the block decomposition proposed by user1551. I proposed another solution that I can detail if you are interested.