determinant inequality $\det(A^2+AB+B^2)\geq\det(AB-BA)$ 
 $A,B$ are two $2\times 2$ real matrices, then 

$$\det(A^2+AB+B^2)\geq\det(AB-BA)$$



The inequality is  equivalent to the following problem: Let $X=A+\dfrac{B}{2},Y=-\dfrac{B}{2}$ 
$$\det[(X-Y)(X+Y)-2(X^2+Y^2)]≥4\det(XY-YX)$$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=353&t=588819

For any two $2\times2$ matrices $A$ and $B$, the following identity holds:
$\renewcommand{\tr}{\operatorname{tr}}$ 
$\renewcommand{\adj}{\operatorname{adj}}$
$$
\det(X+Y) \equiv \det(X) + \det(Y) + \tr(X\adj(Y)).\tag{$\ast$}
$$
Therefore,
\begin{align}
\det(AB-BA)
&=2\det(AB) + \tr(AB\adj(-BA))\\
&=2\det(AB) - \tr(AB\adj(A)\adj(B)).
\end{align}
Write $t=\tr(A\adj(B))=\tr(B\adj(A))$. Then
\begin{align}
&\det(A^2+AB+B^2)\\
=&\det((A+B)^2 - BA)\\
=&\det((A+B)^2) + \det(-BA) + \tr((A+B)^2\adj(-BA))\quad\text{ by } (\ast)\\
=&\det(A+B)^2 + \det(AB) - \tr((A^2+B^2+AB+BA) \adj(A)\adj(B))\\
=&\det(A+B)^2 + \det(AB) - (\det(A)+\det(B))t - \tr(AB \adj(A)\adj(B)) - 2\det(AB)\\
=&\left(\det(A)+\det(B)+t\right)^2 - 3\det(AB) - (\det(A)+\det(B))t + \det(AB-BA)\\
=&\left(t + \frac{\det(A)+\det(B)}2\right)^2 + \frac34\left(\det(A)-\det(B)\right)^2 + \det(AB-BA)\\
\ge&\det(AB-BA).
\end{align}
