If the eigenvalues of a $2\times2$ matrix $A$ have the same $p$-adic absolute value, must $\left|\operatorname{tr}(A)^2\right|\le |\det(A)|$?

Question

Let $A$ be an invertible $2\times2$ matrix over a complete nonarchimedean field $k$ - e.g., $k = \mathbb{Q}_p$.

Suppose its eigenvalues have the same absolute value (in $\overline{k}$). Must we have: $$\left|\operatorname{tr}(A)\right|^2\le \det(A)?$$

(It's certainly true if we replace the right hand side with $4\det(A)$)

Answer 1

Suppose that the eigenvalues are $\lambda_1$ and $\lambda_2$. Then we have $$ |\mathrm{tr}(A)|=|\lambda_1+\lambda_2|\leq \max\{|\lambda_1|,|\lambda_2|\}=|\lambda_1|$$ since by hypothesis $|\lambda_1|=|\lambda_2|$, hence $$ |\mathrm{tr}(A)|^2\leq |\lambda_1|^2=|\lambda_1||\lambda_2|=|\lambda_1\lambda_2|=|\det(A)|$$ as desired.