Prove that $\det (A^n + B^n) \geq 0 $ Let $A$ and $B$ from $M_n(\mathbb{R})$ with $AB=BA$ and $\det(A+B) \geq 0$
Prove that $ \forall n \in \mathbb{N^{*}} $, $\det (A^n + B^n) \geq 0 $

I tried to use induction but I got stuck. 
 A: HINT: 
You could use complex numbers. Consider the factorization
$$(X^n + 1) = \prod_{\omega^n +1 = 0}( X- \omega)$$
The roots of the equation $X^n+1 = 0$ are complex, distinct, and conjugate in pairs, if $n$ is even, or $-1$, and $\frac{n-1}{2}$ pairs of conjugate roots, if $n$ is odd. 
Since $A$, $B$ commute, we can write the equality in $M_n(\mathbb{C})$
$$A^n + B^n = \prod_{\omega^n + 1 = 0} (A - \omega B)$$
For $n$ even we get 
$$A^n + B^n = \prod_{\omega^n + 1= 0, \Re \omega >0}( A - \omega B) (A - \bar \omega B)$$ while for $n$ odd we get 
$$A^n + B^n = (A+B) \cdot \prod_{\omega^n + 1= 0, \Re \omega >0}( A - \omega B) (A - \bar \omega B)$$
Now take determinants and notice that 
$$\det(A- \omega B) \det (A- \bar \omega B) = |\det(A- \omega B)|^2 \ge 0$$
A: Ok, I really like the answer by orangeskid.
However, I want to understand the solution outlined by Giuseppe Negro. 
As he already noted, commuting diagonalizable matrices are simultaneously diagonalizable. So let us assume that $A=diag(a_1, a_2, \ldots a_n)$ and  $B=diag(b_1, b_2, \ldots b_n)$. Then $$\det(A+B)=\prod_{i=1}^n (a_i+b_i)$$ and $$\det(A^k+B^k)=\prod_{i=1}^n (a_i^k+b_i^k)$$
So we have to prove that $\prod_{i=1}^n (a_i+b_i) \geq0 $ implies $\prod_{i=1}^n (a_i^k+b_i^k) \geq 0$.
If $n$ is even this is true for any $a_i$ and $b_i$ even without the assumption since any even power is non-negative. 
If $n$ is odd then consider two cases. If $a_i+b_i =0$ for some k, then $a_i^k+b_i^k=0$ as well, so the implication is true. Let us consider case when $a_i+b_i \neq 0$ for any $i$. It is suffice to prove that $a_i+b_i$ have the same sign as $a_i^k+b_i^k$. If $a_i$ and $b_i$ have the same sign it is clear. If they have different signs then the sign of a sum is determined by maximum absolute value of a summands.
Ok, how should we continue to get full solution?
I have some very vague ideas about density, Zariski topology, and the fact that value of continuous (and thus polynomial) function is determined by its values on a dense subset, but I do not see if it is a good path. 
First of all, I can prove that diagonalisable matrixes are dense it $M_n(\mathbb{C})$, but I can not do it for $M_n(\mathbb{R})$. Is it even true for $M_n(\mathbb{R})$?
Second, we have the inequality and not the polynomial identity.
So please help me to finish the solution, I hope it will be useful for learning community.
Thanks a lot!!!
Update
I am able to prove that diagonalisable matrixes are not dense it $M_n(\mathbb{R})$, so now this approach does not look promising, so please post your ideas!
