How to show the following problem regarding determinant Let $n \in \mathbb{N},n \ge 2,$ and $A,B \in \mathcal{M}_n(\mathbb{R}).$ Prove that there exists a complex number $z,$ such that $|z|=1$ and $$\Re \left( {\det(A+zB)} \right) \ge \det(A)+\det(B),$$where $\Re(w)$ is the real part of the complex number $w$.
We know that $f(z)=\det(A+zB)$  is a polynomial in $z$. Can we use this idea?
 A: Incomplete proof
Assume first that both $A$ and $B$ are singular matrices. We have to prove that $\mathfrak R(f(z))\ge 0$ for some $z \in \partial D$ where $\partial D$ is the unit complex circle.
$f(z)$ is a complex polynomial whose integral on $\partial D$ vanishes. If $\mathfrak R(f(z))$ is always vanishing on $\partial D$, we’re done. If not it must take positive and negative values (if the integral of a non always vanishing continuous function is equal to zero, the function must take positive and negative values). Therefore we’re also done.
Now, removing our initial assumption, suppose for example that $\det B \neq 0$. We have to prove that
$$\mathfrak R(\det(C+zI)) \ge \det(C)+1$$ for some $z \in \partial D$ where $C =B^{-1}A$.
As $f(z) = \det(C+zI)=z^n +a_{n-1} z^{n-1} +\dots +a_1 z+\det(C),$ we’re left to prove that any real trigonometric polynomial 
$$\cos n\theta + a_{n-1} \cos (n-1)\theta + \dots a_1 \cos \theta -1$$
takes at least a non negative value for $\theta \in \mathbb R$.
and here I don’t know how to follow on...
