How do I guarantee that the extrema found using Lagrange Multipliers are in fact max or min? I've been looking at the forum and this particular question almost answered what I would want to know. The question, nevertheless, is a little different from what I would like to ask.
I'm given a simple Lagrange Multiplier problem:

Miminize $f(x,y,z) = x^2 + y^2 + z^2$ given the constraint $x+y+z = 1$ 

(Larson 10th edition)
My answer (checked on the end of the book) was

$x = y = z = \frac{1}{3} \qquad f\left( \frac{1}{3} , \frac{1}{3} , \frac{1}{3} \right) = \frac{1}{3}$

But the real question (at least to me) is: how do I guarantee that it is in fact a minimum? 
Should I
1. Use intuition
2. Use a hessian matrix (is this case, how exactly for this example)?
Also:
3. Why do books do not mention this (predictable) question?
Thank you.
 A: The Lagrange Multiplier algorithm will give you all the critical points. So, if there is a minimum, it will be in your list and it will be the one(s) that produces the smallest value of $f$. 
So the only relevant question is whether a minimum exists at all. And, in this case and many you know a minimum exists because you can restrict the domain so that it is compact and at least one of the coordinates is greater than one (say)
 in absolute value (for example the set of those $x,y,z$ with $x+y+z=1$ and $x^2+y^2+z^2\leq 3$). As the function $f$ is continuous and applies to a compact set, it achieves its max and its min. 
Note, though, that in order to apply the above you need to consider the edges of the domain as critical points. For instance, treating your example as I mention above the equations that Lagrange Multipliers give you will not express the maximum, because it occurs on the edge (somewhere on the circle). 
As Ted Shifrin mentioned, this goes beyond the typical Several Variables course. 
