Given $A \in M_n(\mathbb{R})$ with $A+A^t=2I_n$ prove that $\det(A)\geq 1$.
This is what I tried to do:
$$A+A^t-2I_n=0 \Rightarrow \det(A+A^t-2I_n)=0$$
and then I considered $f:\mathbb{C} \rightarrow \mathbb{C}$ with $f(x)= \det(A+A^t+xI_n)$.
It is known that $f(x)=x^2+\mathrm{Tr}(A+A^t)x+ \det(A+A^t)$.
So this means $f(-2)=0$ but I don't know what to do next.. Thanks in advance!