Efficient method for minimizing $f = x^4 + 2y^4 + 3z^4$ over the unit sphere. I'm helping a student prepare for a final, and they've been provided with a set of sample problems, with approximate points that the problem would be worth. A typical long-answer, show-all-your-work problem is worth 10pts. Answer-only problems are 5pts. 
The following question is given as an answer-only, 5pt question:

Minimize $f = x^4 + 2y^4 + 3z^4$ over the unit sphere.

But I cannot find an approach to solving it that seems on par with the difficulty of the problem, implied by its context.
I tried the following methods: 


*

*I started in on this using the method of Lagrange multipliers, but the number of sub-cases is unwieldy, and would clearly take more time (not to mention writing space) than makes sense, given the context. 

*I then briefly considered using Lagrange in spherical coordinates, but it is quickly apparent that this too is not a more effective method. 

*I also considered taking the constraint and subbing into the original function, e.g. $x^2 = 1 - (y^2+z^2)$, which yields $$f = 1- 2(y^2 + z^2) + 2y^2z^2 + 3y^4 + 4z^4 $$
The critical point analysis is manageable, especially if one starts with this method (but I don't see how one would know this a priori). 

*Finally, I tried taking a substitution $u = x^2$, $v = y^2$ and $w = z^2$, this gives the system $$\begin{cases}f =  u^2 + 2v^2 + 3w^2 \\[5pt] u + v + w = 1 \end{cases}$$ which is extremely simple to solve, using Lagrange multipliers, and it gives the correct value for the minimum; however, I feel deeply uncomfortable with this substitution: I simply don't know under what conditions such a substitution is justified...
I feel like I'm circumambulating some theory that would simplify this entire question, but I can't put my finger on it. 
I apologize that this isn't a more pointed question -- but I think it's a case of "Knowing the question is knowing the answer."
Any help would be greatly appreciated! Thanks in advance for your time and help!
 A: The direct approach of $(4x^3,8y^3,12z^3)=\lambda(2x,2y,2z)$ seems straightforward.
However your method 4 is a very good one.It is often a good idea with Lagrange multipliers to choose the variables carefully.
A: This is a response to the following question from the OP.
Do you know when / if / how to know whether the choice of variables will affect our solution? I've never seen this done, I just tried it and saw that it worked... 
Consider your change of variable and ask yourself a simple question.
Are the values attained by $x^4 + 2y^4 + 3z^4$ subject to $x^2 + y^2 + z^2 = 1 $ precisely the same as the values of $u^2 + 2v^2 + 3w^2$ subject to $u+v+w = 1, u,v,w\ge 0 $?
If $x^4 + 2y^4 + 3z^4=A$ for $x^2 + y^2 + z^2 = 1 $ then for $u=x^2$ etc we have $u^2 + 2v^2 + 3w^2=A$ with $u+v+w = 1$. The converse is just as easy.
So the values are the same and therefore the minimum values of the two problems are the same, irrespective of any special features of the change of variable. There is therefore no reason to think further about such a change of variable and especially not if you are in an exam where you are expected to give a multiple choice answer!
Also it should be noted that there are no special concerns here about critical values. Since we have seen that the two problems are fully equivalent then there could only be genuine concern about the modified problem if there was something wrong in general with the method of applying Lagrange multipliers.
For your problem you need to look at critical values just as you would for any Lagrange multiplier problem. In the case of your modified problem the values of $1$ for any variable means that the other variables are zero so you can just deal with those cases. Since interchanging the values of $u,v,w$ does not affect the constraint then we clearly only need consider either $w=0$ or $v=w=0$ and there is little difficulty in dealing with these two cases (or quickly dismissing them in a multiple choice exam).    
A: Observe that $f$ is minimum when the two surfaces 
$$\begin{cases}f =  x^4 + 2y^4 + 3z^4 \\[5pt] x^2 + y^2 + z^2 = 1 \end{cases}$$
are tangential to each other. So, match their normal vectors,
$$(\frac{x}{z},\frac{y}{z},1) = (\frac{x^3}{3z^3},\frac{2y^3}{3z^3},1)
\implies x^2 = 2y^2 = 3z^2$$
for the constraint on the points. Then, plug it into the plane equation to get the tangential points 
$$(x^2,y^2,z^2)=(\frac6{11},\frac3{11},\frac2{11})\tag 1$$
Thus, the mimimum $f$ is
$$f_m = \left(\frac6{11}\right)^2 + 2\left(\frac3{11}\right)^2 + 3\left(\frac2{11}\right)^2=\frac6{11}$$
Note that there are multiple points given by (1) that admit the same minimum value.
A: $$\begin{array}{ll} \text{minimize} & \overbrace{x^4 + 2 y^4 + 3 z^4}^{=: f}\\ \text{subject to} & x^2 + y^2 + z^2 = 1\end{array}$$
Introducing Lagrange multiplier $\mu$, we eventually obtain the following system of four polynomial equations in unknowns $x$, $y$, $z$ and $\mu$
$$\begin{aligned} x \left( 2 x^2 - \mu \right) &= 0\\ y \left( 4 y^2 - \mu \right) &= 0\\ z \left(6 z^2 - \mu \right) &= 0\\ x^2 + y^2 + z^2 &= 1\end{aligned}$$
and, hence,
$$\begin{aligned} x = 0 \lor x^2 = \frac{\mu}{2}\\ y = 0 \lor y^2 = \frac{\mu}{4}\\ z = 0 \lor z^2 = \frac{\mu}{6}\\ x^2 + y^2 + z^2 = 1\end{aligned}$$
Let us introduce $\color{blue}{\beta_x, \beta_y, \beta_z \in \{0,1\}}$ and write
$$x^2 = \beta_x \frac{\mu}{2}, \qquad y^2 = \beta_y \frac{\mu}{4}, \qquad z^2 = \beta_z \frac{\mu}{6}$$
Obviously, $\beta_x = \beta_y = \beta_z = 0$ is not admissible, as the origin is not on the unit sphere. Hence, we are left with $2^3 - 1 = 7$ cases. From the equality constraint, we obtain
$$\mu = \frac{12}{6 \beta_x + 3 \beta_y + 2 \beta_z}$$
and, evaluating the objective function $f$,
$$f = \left(\frac{6 \beta_x + 3 \beta_y + 2 \beta_z}{24}\right) \mu^2 = \frac{6}{6 \beta_x + 3 \beta_y + 2 \beta_z}$$
and, thus, the minimum is attained when $\color{blue}{\beta_x = \beta_y = \beta_z = 1}$. The minimum is $f_{\min} = \frac{6}{11}$.
