Determinant inequality $ \det(A^2+B^2+(A-B)^2)\ge 3\det(AB-BA) $ $A$ and $B$ are two  $2\times2$ reals matrices. then
$$ \det \Big(A^2+B^2+(A-B)^2\Big)\ge 3\det\left(AB-BA\right).$$
Well, it is seems interesting, but it is really hard to get started.
Thank you very much!
 A: As shown in Omran Kouba's answer, if we put $T=\sqrt{3}(A-B)$ and $U=A+B$, the inequality in question can be rewritten as
$$
\det(T^2+U^2)\ge\det(TU-UT).
$$
Note that For any two $2\times2$ matrices $A$ and $B$, the following identities hold:
$\renewcommand{\tr}{\operatorname{tr}}$ 
$\renewcommand{\adj}{\operatorname{adj}}$
\begin{align}
\det(A+B) &\equiv \det(A) + \det(B) + \tr(A\adj(B))\tag{1}\\
\tr(A\adj(B)) &\equiv \tr(B\adj(A)).\tag{2}
\end{align}
Let $t=\tr(T\adj(U))=\tr(U\adj(T))$. Put $(A,B)=(T,U),\ (T,-U)$ and $\left(TU-UT,\ U^2-T^2\right)$ in $(1)$, we obtain
\begin{cases}
\det(T-U) &= \det(T)+\det(U)-t,\\
\det(T+U) &= \det(T)+\det(U)+t,\\
\det((T-U)\det(T+U)) &= \det(TU-UT) + \det(T^2-U^2).
\end{cases}
Therefore
$$
-\det(T^2-U^2) = \det(TU-UT) + t^2 - (\det(T)+\det(U))^2.\tag{3}
$$
Now, put $(A,B)=(T^2,U^2)$ and also $(T^2,-U^2)$ in $(1)$ and sum up, we get
$$
\det(T^2+U^2) + \det(T^2-U^2) = 2\det(T)^2+2\det(U)^2.
$$
Thus
\begin{align}
&\det(T^2+U^2)\\
=& 2\det(T^2)+2\det(U^2) - \det(T^2-U^2)\\
=& 2\det(T^2)+2\det(U^2) + \det(TU-UT) + t^2 - (\det(T)+\det(U))^2\quad \text{ by } (3)\\
=& \det(TU-UT) + (\det(T)-\det(U))^2 + t^2\\
\ge& \det(TU-UT).
\end{align}
A: We will use the following Lemmas:

${\bf Lemma~1.}$ Let $R$, $S$ be two real $2\times 2$ matrices. Then
  $$ \Re(\det(R+iS))=\det(R )-\det(S).$$

$Proof.$ Indeed, if $R_1,R_2$ are the column vectors of $R$, and 
$S_1,S_2$ are the column vectors of $S$,  then using the bi-linearity of the determinant we have
$$\eqalign{
\det(R+iS)&=\det(R_1+iS_1,R_2+iS_2)\cr
&=\det(R)-\det(S)+i(\det(R_1,S_2)+\det(R_2,S_1)).}
$$
and Lemma 1  follows. $\qquad\square$

${\bf Lemma~2.}$ Let $T$, $U$ be two real $2\times 2$ matrices. Then
  $$ \vert \det(T+iU) \vert^2=\det(T^2+U^2 )-\det(TU-UT).$$
  In particular, 
  $$\det(TU-UT)\leq \det(T^2+U^2 ).\tag{$*$}$$

$Proof.$ Indeed, 
$$\eqalign{
\vert \det(T+iU) \vert^2&=\det(T-iU)\det(T+iU)\cr&=\det((T-iU) (T+iU))
\cr&=\det(T^2+U^2+i(TU-UT))
}
$$
Now, we apply Lemma 1. with $R=T^2+U^2$ and $S=TU-UT$.$\qquad\square$
Let us come to the proposed question. We will apply $(*)$ with
$$T=\sqrt{3}(A-B),\quad U= A+B $$
We check easily that
$$\eqalign{
TU-UT&=2\sqrt{3}(AB-BA)\cr
T^2+U^2&=2(A^2+B^2+(A-B)^2).}
$$
From this the proposed inequality follows.
