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
 A: $\square$ Since for any $2\times 2$ matrix $M$ one has
$$2\operatorname{det}M=\left(\operatorname{Tr}M\right)^2-\operatorname{Tr}M^2,\tag{1}$$
the inequality we want to prove is equivalent to
\begin{align}\left(\operatorname{Tr}(A^2+B^2+AB)\right)^2&\geq \operatorname{Tr}\left(\left(A^2+B^2+AB\right)^2-\left(AB-BA\right)^2\right)=\\
&=\operatorname{Tr}\left(A^4+B^4-ABAB+2A^3B+2AB^3+4A^2B^2\right). \tag{2}\end{align}
On the other hand, using (1) to rewrite the inequality proved here (mentioned in the comments) and replacing therein $B\rightarrow -B$, we obtain exactly the same inequality (2). $\blacksquare$
A: As invertible matrices are dense in the matrix space, we may assume that $A$ is invertible. Left- and right- multiply both sides by $\det(A^{-1})$, the inequality in question becomes
$$\det(I + BA^{-1} + A^{-1}B\,BA^{-1}) \ge \det(BA^{-1} - A^{-1}B).\tag{1}$$
So, it suffices to prove that
$$\det(I + X + YX) \ge \det(X-Y)\tag{2}$$
for any invertible $X,Y$ that are similar to each other. Let $X=kI+X_0$ and $Y=kI+Y_0$, where $X_0$ and $Y_0$ are two nilpotent matrices. Then $(2)$ can be further rewritten as
$$\det\left((1+k+k^2)I + (k+1)X_0 + kY_0 + Y_0X_0\right) \ge \det(X_0-Y_0).\tag{3}$$
If $X_0=Y_0=0$, the inequality is trivial. Suppose $X_0, Y_0$ are nonzero. So, by a change of basis, we may assume that
$$
Y_0=\pmatrix{0&1\\ 0&0},\ X_0=\pmatrix{p&-q\\ r&-p}$$
with $qr=p^2$. If one calculates both sides of $(3)$ directly using the entries of $X_0$ and $Y_0$, one will find that the LHS is equal to $(1+k+k^2)^2 + r$ and the RHS is $r$. Now the inequality follows immediately.
A: 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}
