<s>[INST] Global maximum and minimum of $f(x,y,z)=xyz$ with the constraint $x^2+2y^2+3z^2=6$ with Lagrange multipliers? The global maximum and the global minimum of the function $f(x,y,z)=xyz$ with the constraint $x^2+2y^2+3z^2=6$ can be found using Lagrange multipliers.
$\nabla f = \lambda \nabla g$
$g(x,y,z)=x^2+2y^2+3z^2-6=0$
This results into:
$$yz = \lambda 2x$$
$$xz = \lambda 4y$$
$$xy=\lambda6z $$
When looking at the equation $x^2+2y^2+3z^2-6=0$, it can be seen that there are six solutions $(x,y,z)$ with two coordinates equal to zero: $(0,0,\pm \sqrt 2 ),  (0, \pm \sqrt 3, 0 ) ,  (\pm \sqrt 6, 0, 0 ) $
But I got stuck here. Could someone help me with this?
 [/INST] I think you may have confused yourself in computing and interpreting your result.  
First, in solving the set of "Lagrange equations" you found, you can follow the suggestions of Mann or KittyL, or try it this way.  We may go ahead and solve each equation for the multiplier $ \ \lambda \ $ thus,
$$ \lambda \ = \ \frac{yz}{2x} \ = \ \frac{xz}{4y} \ = \ \frac{xy}{6z} \ \ . $$
This can be read as three equations, each involving a pair of ratios (one of the equations is redundant).  Since we may neglect the possibility of any of the variables having the value zero, as this would automatically produce zero for the value of $ \ f \ $ , we may solve any two of these equations to find, say,
$$ \frac{yz}{2x} \ = \ \frac{xz}{4y} \ \ \Rightarrow \ \ \frac{y}{2x} \ = \ \frac{x}{4y} \ \ \Rightarrow \ \ 2 x^2 \ = \ 4 y^2 \ \ \text{or} \ \ 2 y^2 \ = \ x^2 \ \ ,  $$
$$ \frac{yz}{2x} \ = \ \frac{xy}{6z} \ \ \Rightarrow \ \ \frac{z}{2x} \ = \ \frac{x}{6z} \ \ \Rightarrow \ \ 2 x^2 \ = \ 6 z^2 \ \ \text{or} \ \ 3 z^2 \ = \ x^2 \ \ ,  $$
with the remaining ratio equation telling us nothing new. [We encounter similar systems of non-linear equations for many "rectangular box" volume optimization problems, for which we may also discard zero variable values.]
Inserting these into the constraint equation produces
$$ x^2 \ + \ 2y^2 \ + \ 3z^2 \ = \ 6 \ \ \Rightarrow \ \ 3 \ x^2 \ = \ 6 \ \ \Rightarrow \ \ x \ = \ \pm \sqrt{2} \ \ ,  $$
$$  y^2 \ = \ \frac{1}{2} \ x^2 \ = \ 1 \ \Rightarrow \ \ y \ = \ \pm 1 \ \ , \ \   z^2 \ = \ \frac{1}{3} \ x^2 \ = \ \frac{2}{3} \ \Rightarrow \ \ z \ = \ \pm \frac{\sqrt{6}}{3} \ \ . $$
The solution points  lie on the surface of a triaxial ellipsoid centered on the origin.  Because the function we are extremizing, $ \ f(x, \ y, \ z) \ = \ x \ y \ z \ $ , has an anti-symmetry about the origin and the constraint surface is symmetric about the origin, we may expect that the extrema will be arranged with such symmetry.  There are in fact eight extremal points, one in each octant of $ \ \mathbb{R}^3 \ $ ; these form four pairs on lines through the origin .  The absolute value of the function is the same for all eight points and the "partners" will have values with opposite signs.  
We can then group the solution points as
maximal value :  $$ \ \frac{2 \sqrt{3}}{3} \ \ \text{at} \ \ ( \ \sqrt{2} \ , \ 1 \ , \ \frac{\sqrt{6}}{3} \ ) \ \ , \ \ ( \ \sqrt{2} \ , \ -1 \ , \ -\frac{\sqrt{6}}{3} \ ) \ \ ,( \ -\sqrt{2} \ , \ -1 \ , \ \frac{\sqrt{6}}{3} \ ) \ \ ( \ -\sqrt{2} \ , \ 1 \ , \ -\frac{\sqrt{6}}{3} \ ) \ \ ;  $$
minimal value :  $$ \ -\frac{2 \sqrt{3}}{3} \ \ \text{at} \ \ ( \ \sqrt{2} \ , \ -1 \ , \ \frac{\sqrt{6}}{3} \ ) \ \ , \ \ ( \ \sqrt{2} \ , \ 1 \ , \ -\frac{\sqrt{6}}{3} \ ) \ \ ,( \ -\sqrt{2} \ , \ 1 \ , \ \frac{\sqrt{6}}{3} \ ) \ \ ( \ -\sqrt{2} \ , \ -1 \ , \ -\frac{\sqrt{6}}{3} \ ) \ \ .  $$
These are confirmed by user64494's Maple computation, although we may discard the results with a function value of zero.
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