An inequality with determinant Let $A$ be an invertible $n \times n$ matrix such that all the elements of $A$ and $A^{-1}$ are integers. Prove that if all the eigenvalues of $A$ are real numbers then: 
$$|\det(A + A^{-1})| \ge 2^n$$ 
I've already known that if $A$ is a diagonalizable matrix with eigenvalues $\lambda_1$, $\lambda_2$,... $\lambda_n$ then we can prove that $|\det(A + A^{-1})| = |(\lambda_1 + \lambda_1^{-1})...(\lambda_n + \lambda_n^{-1})| \ge 2^n$ but I still cannot solve the problem, could somebody please give me some help? Thank you.
 A: $A$ may not be diagonalizable, but it is triangularizable over $\mathbb C$. Since you know that all the eigenvalues of $A$ are real numbers, you get the same eigenvalues when considering $A$ as a matrix with complex entries, thus $A$ is similar to some $$ T=\left(
    \begin{array}{ccccc}
    \lambda_1                                   \\
      & \lambda_2             &   & \huge *\\
      &               & \ddots                \\
      & \large 0 &   & \lambda_{n-1}            \\
      &               &   &   & \lambda_n
    \end{array}
    \right)$$
that is $A=PTP^{-1}$ where $P\in Gl_n(\mathbb C)$. $T^{-1}$ is upper triangular as well, with the form
$$ T^{-1}=\left(
    \begin{array}{ccccc}
    \lambda_1^{-1}                                    \\
      & \lambda_2 ^{-1}            &   & \huge **\\
      &               & \ddots                \\
      & \large 0 &   & \lambda_{n-1} ^{-1}           \\
      &               &   &   & \lambda_n^{-1}
    \end{array}
    \right)$$
so that $A^{-1}=PT^{-1}P^{-1}$, which yields $|\det(A + A^{-1})| = |(\lambda_1 + \lambda_1^{-1})...(\lambda_n + \lambda_n^{-1})|$.
A: Recall that $\det(A)=\prod_{i=1}^n \lambda_i$, where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$. 
Hence, $\det(A^{-1})=\dfrac{1}{\prod_{i=1}^n \lambda_i}$ and $\det(A^2+Id)=\det(A+iId)\det(A-iId)=\prod_{j=1}^n (\lambda_j+i)\prod_{j=1}^n (\lambda_j-i)=\prod_{j=1}^n (\lambda_j^2+1)$.
Since $\lambda_i\in\mathbb{R}$ then $\lambda_i$ and $\lambda_i^{-1}$ have the same sign. So $|\lambda_i+\lambda_i^{-1}|=|\lambda_i|+|\lambda_i^{-1}|$.
Since $|\lambda_i|^2-2|\lambda_i|+1\geq 0$ then $|\lambda_i|^2+1\geq 2|\lambda_i|$. So $|\lambda_i|+|\lambda_i^{-1}|\geq 2$.
Finally, $|\det(A+A^{-1})|=|\det(A^{-1})\det(A^2+Id)|=\left|\prod_{i=1}^n \dfrac{\lambda_i^2+1}{\lambda_i}\right|=\prod_{i=1}^n \left|\lambda_i+\dfrac{1}{\lambda_i}\right|\geq 2^n$.
A: \begin{eqnarray*}
\lambda+ \frac{1}{\lambda} =\left( \sqrt{\lambda} - \frac{1}{\sqrt{\lambda}} \right)^2+2 \geq 2.
\end{eqnarray*}
