Let $A, B\in M_n(R)$ and $AB=O_n$ (where $O_n$ means the zero matrix). Prove that $$\det(A^2+B^2+A+B+I_n) \ge 0$$
I tried to write this as the product of two complex conjugates, and then the determinant would be $\ge 0$,but I can't manage to do this.
EDIT: the problem has two tasks, and the first was to prove $\det(A^2+A+I_n)\ge 0$, which I solved by writing $\det(A^2+A+I_n)=\det(A+\epsilon I_n) \cdot \det(A+\epsilon^2 I_n) \ge 0$,where $\epsilon$ is a complex third root of unity.
Best Answer
Here is a simple proof that I have just found.
We have proved that $\det(X^2+X+I_n) \ge 0$, $\forall X \in M_n(\mathbb R) $.
Also, $(A^2+A+I_n)(B^2+B+I_n)=A^2+B^2+A+B+I_n$ and by taking the determinant of this relation we are done.