Disclaimer: In the definition (Stewart Calculus, 7E): "Method of Lagrange Multipliers" part (b)- Evaluate $f$ at all extreme points $(x,y,z)$ from step a. The largest of these values is the maximum and the smallest of these values are the minimum.
It is impossible to infer whether an extreme point is a maximum or minimum from this definition when there is only one extreme point in the constraint. Can it be either proven or inferred that it is a maximum or minimum value using another method? Wolfram Alpha Lagrange Calculator said it was a minima (this is an even number exercise, ie no answer in back of book).
Second Disclaimer: I will show my work as to how I only got one value for verification, otherwise if not verified and there is an error the question could be ended there depending on the nature of the error. Then I would like you to tell me how/if it can be proven or inferred that a extreme value in Lagrange's method is a max or min when Stewart's definition doesn't apply and there is only one point.
Function: $f(x,y,z) = x^2 + y^2 + z^2$
Constraint: $x + y + z = 12$
$\nabla f$ = $\lambda \nabla g $:
$2x = \lambda$
$2y = \lambda$
$2z = \lambda$
Multiplying Equation 1 by yz, Equation 2 by xz, Equation 3 by xy. Gives the following:
$2xyz = \lambda yz$
$2xyz = \lambda xz$
$2xyz = \lambda xy$
$\lambda yz = \lambda xz = \lambda xy \to x=y=z$
We will say that $x=y=z=a$ so that plugging into the constraint gives:
$3a = 12 \to a = 4 \to x=y=z=4$
We have our only extreme value at $f(4,4,4)= 48
How is this a minimum?