Finding the distance from the origin to the surface $xy^2 z^4 = 32$ using the method of Lagrange Multipliers Problem: Find the distance from the origin to the surface $xy^2z^4 = 32$.
Attempt: The Lagrange equation for this problem is $L(x,y,z, \lambda) = x^2 + y^2 + z^2 + \lambda (xy^2 z^4 - 32)$. Setting the first partials to zero we have \begin{align*} \frac{\partial L}{\partial x} &= 2x + \lambda y^2 z^4 = 0 \qquad (1)  \\ \frac{\partial L}{\partial y} &= 2y + 2 \lambda x y z^4 = 0 \qquad (2) \\ \frac{\partial L}{\partial z} &= 2z + 4 \lambda x y^2 z^3 = 0 \qquad (3) \\ \frac{\partial L}{\partial \lambda} &= xy^2 z^4 - 32 = 0 \qquad (4)
\end{align*}
Now I'm having a hard time solving this system for $x,y$ and $z$. Here is what I did so far. From $(1)$ and $(2)$ we get \begin{align*} \frac{2x}{y^2 z^4} = - \lambda \qquad \text{and} \qquad \frac{1}{xz^4} = - \lambda \end{align*} Thus $\frac{2x}{y^2 z^4}  = \frac{1}{xz^4} $ or $y^2 = 2x^2$ after simplification. Also, from $(2)$ and $(3)$ we can deduce that \begin{align*} \frac{1}{xz^4} = - \lambda = \frac{2z}{4xy^2 z^3} \end{align*} so that $2y^2 = z^2$ after simplification. Now I used all this and substituted it into $(4)$. This gave me \begin{align*} x(2x^2) (4y^4) - 32 = 0 \end{align*} or (since $y^4 = 4x^4)$ \begin{align*} 8x^3 (4x^4) - 32 = 0 \end{align*} 
This means that $32x^7 - 32 = 0$, so that $x = 1$. Then $y^2 = 2$, so that $y = \pm \sqrt{2}$. Then $z^2 = 4$, so that $z = \pm 2$. So I found the points $(x,y,z) = (1, \sqrt{2}, 2)$ and $(1, - \sqrt{2}, -2)$. They both give me the distance $\sqrt{x^2 + y^2 + z^2} = \sqrt{7}$, so I'm guessing they are equal? Is my reasoning correct?
 A: Your solution is correct, except the last part.  You should get four minimizing points.  (You mistakenly assumed that $y$ and $z$ must be both positive or both negative.)
However, there is a solution without using Lagrange multipliers.  Note by AM-GM that
$$x^2+y^2+z^2=x^2+2\left(\frac{y^2}{2}\right)+4\left(\frac{z^2}{4}\right)\geq 7\sqrt[7]{x^2\left(\frac{y^2}{2}\right)^2\left(\frac{z^2}{4}\right)^4}=7\sqrt[7]{\left(\frac{xy^2z^4}{32}\right)^2}=7\,.$$
The equality holds iff $x^2=\frac{y^2}{2}=\frac{z^2}{4}$, which means $(x,y,z)=\big(1,\pm\sqrt{2},\pm 2\big)$.  Hence, the required distance is $\sqrt{7}$.
A: Another possibility. First, 
$$\tag{$*$}
xy^2z^4 = 32 \Longrightarrow y^2 = \frac{32}{xz^4}.
$$
Notice that we can arbitrarily divide by $x,y$, or $z$ since any point with a coordinate equal to $0$ do not belong to your surface.
Now, consider the squared norm of a generic point $(x,y,z)$ and use $(*)$:
$$
x^2+y^2+z^2 = x^2 + \frac{32}{xz^4} + z^2 =: N(x,z).
$$
We want to minimize $N(x,z)$ over $\mathbb{R}^2$. Thus, looking for zeros of the gradient, we obtain the system
$$
\begin{cases}
0 = \frac{\partial N}{\partial x}(x,z) = 2x - \frac{32}{x^2z^4},\\
0 = \frac{\partial N}{\partial z}(x,z) = 2z - \frac{128}{xz^5},
\end{cases}
$$
that is solved by $x = 1$ and $z = \pm 2$. In particular, the minimum is $7$. 
A: A hint: 
Multiply $(1)$ by $x$, $(2)$ by $y$, and $(3)$ by $z$ and look at the three equations you got.
