In one of my textbooks, there is this inequality for square matrices larger than $2 \times 2$: $$\tag{0} \lambda_{\min} \ge \det(A)$$
This inequality is written in the context of autocorrelation matrices, so I suppose that $A$ is symmetric and positive semi-definite. But even with these constraints, this inequality doesn't seem to be always true, as I can give a counterexample: $$\left[\matrix{2 && 0 && 0 \\ 0 && 2 && 0 \\ 0 && 0 && 2}\right]$$ This matrix has eigenvalues of 2, and determinant of 8, so the inequality is false.
Is this inequality well known for some kind of matrices? Or maybe it is true when all eigenvalues are less than one (I'm just guessing this...)?
$(0)$ is part of the proof for this inequality: $$\tag{1}\frac{\lambda_\max}{\lambda_\min} \le \frac{\mathrm{Tr}(A)}{\det(A)}.$$
And the proof is: $$\tag{2}\lambda_\max \le \mathrm{Tr}(A)$$ and the inequality in question: $$\tag{3} \lambda_{\min} \ge \det(A)$$
$(2)$ is true (I think), but $(3)$ is not always true, as I have given a counterexample. So maybe, $(1)$ isn't true either?