Question. Given a positive-definite $n \times n$ matrix $A = (a_{ij})$ with eigenvalues $$ \lambda_1 \leq \cdots \leq \lambda_n , $$ is there a sharp upper bound for the product $\lambda_2 \cdots \lambda_n$ in terms of the quantity $$ \|A\|_\infty := \max_{1 \leq i, j \leq n} |a_{ij}| ? $$

A classic inequality due to A. Hirsch states that the modulus of an eigenvalue of an $n \times n$ complex matrix $A$ is less than $n \|A\|_\infty$, which implies that $$ |\lambda_2 \cdots \lambda_n| \leq n^{n-1}\|A\|_\infty^{n-1} . $$ However, this seems like a rather rough estimate for positive-definite matrices. I'm interested in any estimate that is substantially better than this.

Motivation. Using Hirsch's inequality one can improve Lemma 4 in page 43 of Siegel's Lectures on Quadratic Forms to yield the following result:

Theorem. If $A = (a_{ij})$ is a positive-definite $n \times n$ matrix, then for every $x \in \mathbb{R}^n$ we have that $$ \frac{\det(A)}{n^{n-1}a_{11} a_{22} \cdots a_{nn}} \sum a_{ii} x_i^2 \leq \sum a_{ij} x_i x_j \leq n \sum a_{ii} x_i^2 . $$

The inequality on the left would be greatly improved if we had the sharp upper bound required in the question. This in turn would yield a better answer to this enclosure problem (see my answer to that question).