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.
Best Answer
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.