An interesting challenge arises in working with the Lagrange-multiplier method in that all but the simplest problems (generally those involving linear functions and linear constraint equations) can easily lead to systems of multi-linear or non-linear equations, which often do not have general methods of solution. I find that one must draw on an assortment of possible approaches in looking for reasonable routes to a solution.
This problem turns out not to be one where we would solve the set of "Lagrange equations" for expressions for $ \ \lambda \ $ , which we could then equate and proceed from there. The system that Daniel Robert-Nicoud presents can be arranged in this fashion, which is suggestive:
$$ \begin{array}{cccc}8 \ x&- \ \lambda \ y& - \ 3\lambda \ z& = 0\\\lambda \ x& \ -6 \ y& + \ 2\lambda \ z& = 0\\ 3\lambda \ x&+ \ 2 \lambda \ y& \ - 10 \ z& = 0\\\end{array} \ \ . $$
We know that the solution to this system will be non-trivial if the determinant of the coefficients is singular. This is reminiscent of looking for eigenvalues, but we are instead solving the determinant polynomial $ \ -12 \ \lambda^3 \ - \ 96 \ \lambda^2 \ + \ 480 \ = \ 0 \ $ for values of the Lagrange-multiplier that will allow the values of the variables to be other than all zeroes. We find $ \ \lambda \ = \ 2, \ -5 \ \pm \sqrt{5} \ $ .
For $ \ \lambda \ = \ 2 \ $ , we obtain the result indicated by KittyL , $ \ x \ = \ y \ = \ z \ $ . For the other values of the multiplier, we find (with the acknowledged help of WolframAlpha)
$$ \lambda \ = \ -5 \ + \ \sqrt{5} \ \ : \ \ y \ = \ \frac{(5 - \sqrt{5}) \ (10 - 3 \sqrt{5})}{10 \ (2 \sqrt 5 - 3)} \ x \ \approx \ 0.61803 \ x \ \ , \ \ $$
$$ z \ = \ -\frac{(5 - \sqrt{5}) \ (4 + \sqrt{5})}{10 \ (2 \sqrt 5 - 3)} \ x \ \approx \ -1.17082 \ x \ \ ; $$
$$ \lambda \ = \ -5 \ - \ \sqrt{5} \ \ : \ \ y \ = \ -\frac{(5 + \sqrt{5}) \ (10 + 3 \sqrt{5})}{10 \ (2 \sqrt 5 + 3)} \ x \ \approx \ -1.61803 \ x \ \ , \ \ $$
$$ z \ = \ \frac{(5 + \sqrt{5}) \ (4 - \sqrt{5})}{10 \ (2 \sqrt 5 + 3)} \ x \ \approx \ 0.17082 \ x \ \ . $$
Inserting the result for $ \ \lambda \ = \ 2 \ $ into the constraint equation produces the simple result $ \ 6 \ x^2 \ = \ 6 \ \ \Rightarrow \ \ x \ = \ y \ = \ z \ \pm 1 \ $ , for which the value of the function at either point is $ \ 4 \ + \ 3 \ + \ 5 \ = \ 12 \ $ .
Is there anything else worth mentioning? If we apply the values of the variables for the $ \ \lambda \ = \ -5 \ + \ \sqrt{5} \ $ case, the constraint equation yields (and I hope I may be forgiven using the decimal approximations)
$$ 0.61803 \ x^2 \ + \ 2 \ (0.61803) \ (-1.17082) \ x^2 \ + \ 3 \ (-1.17082) \ x^2 \ \approx \ 6 $$
$$ \Rightarrow \ \ -4.34164 \ x^2 \ = \ 6 \ \ , $$
which admits no solution. Finally, the third value $ \ \lambda \ = \ -5 \ - \ \sqrt{5} \ $ leads to
$$ -1.61803 \ x^2 \ + \ 2 \ (-1.61803) \ (0.17082) \ x^2 \ + \ 3 \ (0.17082) \ x^2 \ \approx \ 6 $$
$$ \Rightarrow \ \ -1.65836 \ x^2 \ = \ 6 \ \ . $$
So there is only the one pair of extremal points $ \ (1, \ 1, \ 1) \ $ and $ \ (-1, \ -1, \ -1 ) \ $ , and it is easily checked (even just by trying simple integer triples that solve the constraint equation) that the value found for the function is minimal. Because the function and the constraint equation have symmetry about the origin, it would be expected that extremal solutions of the same value would be found in pairs observing this symmetry.
[I reported this approach because it does prove useful in solving such systems of equations. This one had somewhat complicated values for the multiplier, so it is not simply to work through entirely "by hand". (I also thought I had found another extremum of interest, but in typing this out, I uncovered a sign-error in a late stage of my work...) It is also curious that decimal approximations of the "golden ratio" appear; I've no idea whether this is just coincidental. ]
