Prove that $\det(A^{-1}-A)+\det(A^{-1}+A)\geq 6$ 
Let $A$ be a $3 \times 3$ matrix with real numbers, with $\det(A)=1$ and $\operatorname{Tr}(A)=-1$. Prove that $$\det(A^{-1}-A)+\det(A^{-1}+A)\geq 6.$$

This is supposed to be a 11th grade problem.
Attempts:
I used the formula $\det(A+xB)=\det(A)+ax+bx^{2}+\det(A)x^{3}$ and then
I tried expanding the determinant $\det(A^{-1}-A)=\det(A^{-1})-a+b-\det(A)$, with $\det(A)=\frac{1}{\det A}$, and the same with $\det(A^{-1}+A)=\det(A^{-1})+c+d+\det(A)$ but then I got stuck.
I also used the equation of the matrix $A_{3 \times 3}$:
$A^{3}-\operatorname{Tr}(A)A^{2}+\operatorname{Tr}(A^{*})A-\det(A)I_3=O_3$,
I also know that  $\operatorname{Tr}(A^{*})=\frac{\operatorname{Tr}(A)^{2}-\operatorname{Tr}(A^{2})}{2}$ but I couldn't find $\operatorname{Tr}(A^2)$.
I am missing something or is there another way of solving maybe involving matrix ranks and Sylvester’s Theorem? I could't find a way to apply it. Any help would be appreciated.
 A: Let $p(x)=\det(xI_3-A)$ be the characteristic polynomial of $A$. By multiplying with $\det(A)$, the inequality writes $$-p(1)p(-1)+p(i)p(-i)\ge6\quad(*)$$
Since $\mathrm{Tr}(A)=-1$ and $\det(A)=1$ we get $p(x)=x^3+x^2+ax-1$. Then $(*)$ is equivalent to $(a+1)^2+4+(a-1)^2\ge6$ which is obvious.
A: EDIT: I've rewritten my answer based on a now deleted comment of mine, because I had made an embarrassing mistake in the comment.
The characteristic polynomial of $A$ is $p_A(x)=x^3+x^2+kx-1$.
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We split the question into several cases, depending on the Jordan normal form of $A$:

*

*$A$ has one real eigenvalue $r$, and two distinct complex eigenvalues $z$ and $\overline{z}$.

*$A$ has three distinct real eigenvalues $r$, $s$ and $t$.

*$A$ has two distinct real eigenvalues $r$ and $s$, and furthermore $A$ can be represented as $\begin{bmatrix}r & 1 & 0\\
                                 0 & r & 0\\
                                 0 & 0 & s\end{bmatrix}$.

*$A$ has two distinct real eigenvalues  $r$ and $s$, and furthermore $A$ can be represented as $\begin{bmatrix}r & 0 & 0\\
                                 0 & r & 0\\
                                 0 & 0 & s\end{bmatrix}$.

*$A$ has one real eigenvalue $r$, and furthermore $A$ can be represented as $\begin{bmatrix}r & 1 & 0\\
                                 0 & r & 1\\
                                 0 & 0 & r\end{bmatrix}$ or $\begin{bmatrix}r & 1 & 0\\
                                 0 & r & 0\\
                                 0 & 0 & r\end{bmatrix}$.

*$A=rI$
Note that $\det(A)=1$ forces the eigenvalues to be non-zero.
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In each of cases 1, 2, 3 and 5: it is not hard to check that the vector space is spanned by $v,Av,A^2 v$ for some $v$. In such cases, we can revert to the rational canonical form of $A$, which will be $\begin{bmatrix}0 & 0 & 1\\
                                 1 & 0 & -k\\
                                 0 & 1 & -1\end{bmatrix}$.
Plug that straight into $\det(A^{-1}-A)+\det(A^{-1}+A)$. Here's the Wolfram Alpha link to that computation. It turns out that $\det(A^{-1}-A)+\det(A^{-1}+A)=2k^2+6$.
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In case 4, the constraints $r^2 s=1$, $2r+s=-1$ and $r,s\in\mathbb{R}$ allow only one possibility: $r=-1,s=1$. In this case, $\det(A^{-1}-A)+\det(A^{-1}+A)=8$.
$\\$
Case 6 doesn't fit the constraints on the determinant and trace.
