determinant inequality, $AB=BA$, then $ \det(A^2+B^2)\ge \det(2AB) $

Question

$A$ and $B$ are two $n\times n $ real matrices, $AB=BA$. Can we conclude that

$$ \det \Big(A^2+B^2\Big)\ge \det(2AB) $$

is right?

Well, the inequality is interesting. if $A,B$ are upper triangular matrices, it is obvious right. If $AB\ne BA$, $ \det \Big(A^2+B^2\Big)\ge \det(AB+BA) $ is wrong.

Answer 1

The answer is NO. Take for example $A=I_2$ and

$$ B=\left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right) $$

We then have $B^2=-I_2$, $A^2+B^2=0$ and $2AB=2B$, so ${\sf det}(A^2+B^2)=0$ and ${\sf det}(2AB)=4$.