Extreme value verification using Bordered Hessian matrix Determine the minimum value of $$bcx+cay+abz$$ subject to the conditions $$xyz=abc$$
Here is the solution in the book (using the Lagrange multipliers)
$$F = bcx+cay+abz+k(xyz-abc)$$
After solving it, the minimum value given in the book, comes out at $x=a, y=b, z=c$ 
I was trying to verify whether this answer corresponds to minimum or maximum value using bordered Hessian matrix (let's call it $H$).
$$H = \begin{bmatrix}0&bc&ca&ab\\bc&0&-c&-b\\ac&-c&0&-a\\ab&-b&-a&0\end{bmatrix}$$
$$H = -3a^2b^2c^2$$
$$H < 0$$
This implies $x=a, y=b, z=c$ correspond to maximum value
Where am I going wrong?
 A: Even in the unconstrained case it is not sufficient to only check the determinant of the Hessian, and one needs to calculate the determinant of the minors in order to classify exterma.
In the constrained case things get more complicated, as you are only allowed to move on the constrained surface $xyz=abc$ (in this instance). It means that you need to discard eigenvalues associated with directions out of this surface.
The way to do it is outlined in Wikipedia (in the following $m$ is the number of constraints)

Specifically, sign conditions are imposed on the sequence of leading principal minors (determinants of upper-left-justified sub-matrices) of the bordered Hessian, for which the first $2m$ leading principal minors are neglected... A sufficient condition for a local minimum is that all of these minors have the sign of $\left(–1\right)^{m}$.

Applying this in your case, $m=1$ so we discard the first $2m=2$ minors and we need to calculate only
$$H_{3}=\begin{vmatrix}0&bc&ca\\bc&0&-c\\ac&-c&0\end{vmatrix}=-2abc^{3}<0$$
$$H_{4}=\begin{vmatrix}0&bc&ca&ab\\bc&0&-c&-b\\ac&-c&0&-a\\ab&-b&-a&0\end{vmatrix}=-3a^{2}b^{2}c^{2}<0$$
Both of which has a sign of $\left(-1\right)^{m}=-1$, so this results in a minima.
