Knowing that $A$ and $B$ are $2$ matrices of $2\times2$ with elements on $\mathbb{R}$, we have to prove that $\det(A^2+B^2)\ge\det(AB-BA)$ I tried calculating it directly and got nothing. Any ideas?
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Thanks for the edit. I'm pretty much new to this site and also quite tired right now. – Wolfuryo Nov 28 '18 at 20:34
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Answered in the answer provided by user1551 in https://math.stackexchange.com/questions/756373/determinant-inequality-deta2b2a-b2-ge-3-detab-ba – NoChance Nov 28 '18 at 21:19