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How can I prove :
$$\mathrm{det}(A^2+AB+B^2) \geq (\det(A)-\det(B))^2$$ where $AB=BA$ and $A,B \in \mathbb M_2(R)$?

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    $\begingroup$ Take for e.g. $A=-B$ non singular... $\endgroup$ – Macavity Oct 17 '17 at 14:30
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This is false even under the extra hypothesis that $A$ and $B$ are diagonal matrices (in which case it is automatically true that $AB=BA$). Indeed, if $A=a\operatorname{Id}_2$ and $B=b\operatorname{Id}_2$, then$$\det(A^2+AB+B^2)-\bigl(\det(A)-\det(B)\bigr)^2=ab(2a^2+5 a b+2 b^2).$$Now take, for instance, $a=-3$ and $b=1$. Then, the previous expression is equal to $-15$.

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The inequality is false. E.g. when $$ A=\pmatrix{-1&-2\\ 2&-1},\ B=I, $$ the inequality would wrongly imply that $13\ge16$.

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