How to prove the matrix inequality $\sqrt{\det{(A^2-AB+B^2)}}+\sqrt{\det{(A^2+AB+B^2)}}\ge |\det{(A)}+\det{(B)}|$ Let $A,B\in M_{2}(R)$ be two square matrices so that
$AB=BA$, prove the inequality
$$\sqrt{\det{(A^2-AB+B^2)}}+\sqrt{\det{(A^2+AB+B^2)}}\ge |\det{(A)}
+\det{(B)}|$$
I only know prove this
$$\sqrt{a^2-ab+b^2}+\sqrt{a^2+ab+b^2}\ge |a+b|,a,b\in R$$
because
$$\sqrt{x}+\sqrt{y}\ge\sqrt{x+y}$$
so
$$LHS\ge\sqrt{2(a^2+b^2)}\ge (a+b)$$
But for matrices I can't prove it. Thank you

I'm late to the party and I haven't time to cook up a full solution, but here are some observations.
As $A$ commutes with $B+\epsilon I$ for every real $\epsilon$, by continuity, it suffices to prove the inequality for invertible $B$. Let $X=B^{-1}A$. Divide both sides of the inequality by $|\det(B)|$, the inequality becomes
$$
\sqrt{\det{(X^2-X+I)}}+\sqrt{\det{(X^2+X+I)}}\ge |\det{(X)}+1|.
$$
When $X$ has two real eigenvalues $x,y$, the inequality becomes
$$
\sqrt{(x^2 - x + 1)(y^2 - y + 1)} + \sqrt{(x^2 + x + 1)(y^2 + y + 1)}
\ge |1 + xy|.\tag{1}
$$
When $X$ has a pair of conjugate complex eigenvalues $z$ and $\bar{z}$, the inequality becomes
$$
|z^2 - z + 1| + |z^2 + z + 1| \ge 1 + |z|^2.\tag{2}
$$
Hence the problem boils down to proving $(1)$ and $(2)$.
