Let $\,\lambda_1,...,\lambda_n\,$ be the matrix's eigenvalues (perhaps in some field extension of the original one), which are all positive (of course, it is customary to consider only Hermitian, or symmetric, matrices when defining positive definite), then

$$\det A=\prod_{k=1}^n \lambda_k\,\,\,,\,\,\,\operatorname{tr.}A=\sum_{k=1}^n\lambda_k$$

Thus we're required to prove

$$\prod_{k=1}^n\lambda_k\leq\left(\frac{\sum_{k=1}^n\lambda_k}{n}\right)^n\Longleftrightarrow \sqrt[n]{\prod_{k=1}^n\lambda_k}\leq \,\frac{1}{n}\sum_{k=1}^n\lambda_k$$

which is precisely the AM-GM inequality, as you mentioned.

Upper Bound on Determinant of Matrix in terms of Trace
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