<s>[INST] 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
 [/INST] As user 1551 points out, it suffices to prove the inequality for $AB = BA$ where, $A,B$ are invertible. Since, commuting matrices preserve each others eigenspaces, they are simultaneous diagonalisable, $\exists \, U \in U_2(\mathbb{R})$, such that $U^{-1}AU = D[x_1,y_1] = D_1$ and $U^{-1}BU = D[x_2,y_2] = D_2$.
Thus, $\displaystyle \sqrt{\det{(A^2-AB+B^2)}}+\sqrt{\det{(A^2+AB+B^2)}}\ge |\det{(A)}
+\det{(B)}| \iff \sqrt{\det{(D_1^2-D_1D_2+D_2^2)}}+\sqrt{\det{(D_1^2+D_1D_2+D_2^2)}}\ge |\det{(D_1)}
+\det{(D_2)}|$ 
(since, $|\det(U)| = 1$)
That is, 
$\displaystyle \sqrt{(x_1^2-x_1x_2+x_2^2)(y_1^2-y_1y_2+y_2^2)}+\sqrt{(x_1^2+x_1x_2+x_2^2)(y_1^2+y_1y_2+y_2^2)} \ge |x_1y_1+x_2y_2| \iff \left(\frac{3}{4}(x_1-x_2)^2+\frac{1}{4}(x_1+x_2)^2\right)^{1/2}\left(\frac{3}{4}(y_1-y_2)^2+\frac{1}{4}(y_1+y_2)^2\right)^{1/2} + \left(\frac{1}{4}(x_1-x_2)^2+\frac{3}{4}(x_1+x_2)^2\right)^{1/2}\left(\frac{1}{4}(y_1-y_2)^2+\frac{3}{4}(y_1+y_2)^2\right)^{1/2} \ge_{C.S.} \frac{3}{4}|x_1-x_2||y_1-y_2|+\frac{1}{4}|x_1+x_2||y_1+y_2|+\frac{1}{4}|x_1-x_2||y_1-y_2|+\frac{3}{4}|x_1+x_2||y_1+y_2| \ge |x_1y_1+x_2y_2|$ (by triangle inequality).
</s>