When is it true that $\lambda_{\min} \ge \det(A)$ if $A$ is a symmetric positive semi-definite matrix? 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?
 A: A symmetric positive semi-definite real matrix is a correlation matrix if, and only if, all of its diagonal entries are equal to $1.$ Thus your proposed counterexample fails if it was assumed that it's a correlation matrix.
(I'm going to write a more leisurely answer and post it here tonight or tomorrow.)
A: Since $$\det(A)=\prod \lambda_i$$
we have that for $\lambda_{\min}=0$ the inequality is always true and for $\lambda_{\min}\neq 0$
$$\lambda_{\min} \ge \det(A)\iff \lambda_{\min}\ge\prod \lambda_i \iff 1\ge \frac{\prod \lambda_i}{\lambda_{\min}}$$
A: This is clearly not true for matrices that are symmetric and positive definite with diagonal elements that are one (the typical definition of an autocorrelation matrix.)  Try for example:
>> A=[1 -0.5 0; -0.5 1 -0.4; 0 -0.4 1]
A =

   1.00000  -0.50000   0.00000
  -0.50000   1.00000  -0.40000
   0.00000  -0.40000   1.00000

>> det(A)
ans =  0.59000
>> eig(A)
ans =

   0.35969
   1.00000
   1.64031

Further restricting the matrix to be Toeplitz (as it would be if it were the autocorrelation matrix of a stationary stochastic process) doesn't help:
>> B=[1 -0.5 0; -0.5 1 -0.5; 0 -0.5 1]
B =

   1.00000  -0.50000   0.00000
  -0.50000   1.00000  -0.50000
   0.00000  -0.50000   1.00000

>> eig(B)
ans =

   0.29289
   1.00000
   1.70711

>> det(B)
ans =  0.50000
>>

A: Here's an example of an autocorrelation matrix that doesn't satisfy the proposed inequality (1).   
$\frac{\lambda_{max}}{\lambda_{min}} \leq \frac{\mbox{tr}(A)}{\det(A)} $
>> C=[1 -0.70 0; -0.70 1 -0.70; 0 -0.70 1]
C =

   1.00000  -0.70000   0.00000
  -0.70000   1.00000  -0.70000
   0.00000  -0.70000   1.00000

>> trace(C)/det(C)
ans =  150.00
>> lambda=eig(C)
lambda =

   0.010051
   1.000000
   1.989949

>> lambda(3)/lambda(1)
ans =  197.99

