Romanian Mathematical Olympiad, 11th grade, 2008, problem 1

Showing an inequality involving determinants of two by two matrices. Even though we are dealing with matrices with real entries, we introduce the primitive cube roots of unity to factorize some expressions. We make us of the fact that determinants appearing in our problem can be expressed as some values of the characteristic polynomial of our matrix. This nice problem was given to 11th graders at the Romanian District Olympiad, 2011.

[dariuschitu]

Even though this solution is basically isomorphic to yours (anyway, all solutions ultimately lead to (det(A)=2Tr(A)+1)^2>=0), I think it is a bit cleaner:
Let Spec(A)={ L1, L2 } be the eigenvalues of A. The problem is equivalent to (L1^2 + L1 + 1)(L2^2 + L2 + 1) - 3/4*(1 - L1*L2)^2 >= 0. Factoring, this inequality is just 1/4*(L1*L2 + 2(L1 + L2) + 1)^2 >= 0. But L1*L2=det(A) real, L1+L2=Tr(A) real, so the expression is just 1/4*(det(A)=2Tr(A)+1)^2>=0 which is true because we know we are dealing with reals.