Finding the minimum distance from the origin to the surface $xyz^2=2$ This was an old exam question I was looking at for a friend, although it's been a while since I've done this stuff:
Q. Find the shortest distance from the origin to the surface $xyz^2=2$.
I remembered Lagrange Multipliers being used in that course, so I did the following.
Define $f(x,y,z)=\sqrt{x^2+y^2+z^2}$, which is what we want to minimize, with the constraint $g(x,y,z)=xyz^2=2$.
We get three equations (I'm using $f$ as shorthand to avoid the messy root):


*

*$\frac{x}{f}=\lambda yz^2$ or $\frac{x^2}{f}=2\lambda$

*$\frac{y}{f}=\lambda xz^2$ or $\frac{y^2}{f}=2\lambda$

*$\frac{z}{f}=2\lambda xyz$ or $\frac{z^2}{f}=4\lambda$


We get $y=\pm x$, $z=\pm \sqrt{2} \cdot x$ and with the constraint only $y=x$ will work.
Finally we get $x=\pm 1$ and this gives us $4$ points $\pm(1,1,\sqrt2)$ and $\pm(1,1,-\sqrt2)$.
I haven't really used $\lambda$ though... so it's like I didn't really do it properly.
 A: No one can be zero. So, in that surface, $z^2=2/xy$. Now, by AM-GM
$$x^2+y^2+z^2=x^2+y^2+\frac{2}{xy}\geq2xy +\frac{2}{xy}\geq 2\sqrt{4}=4.$$ Study the conditions for the equality to happen and show that they actually happen for the points you already suspect.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Use $\ds{\tt\mbox{Spherical Coordinates}}$ and
a $\ds{\tt\mbox{Lagrange Multiplier}}$ $\ds{4\mu}$:

\begin{align}
{\cal F}&\equiv r - 4\mu\braces{\bracks{r\sin\pars{\theta}\cos\pars{\phi}}
\bracks{r\sin\pars{\theta}\sin\pars{\phi}}
\bracks{r\cos\pars{\theta}}^{2} -2}
\\[3mm]&=r-4\mu\bracks{r^{4}\sin^{2}\pars{\theta}\cos^{2}\pars{\theta}\sin\pars{\phi}
\cos\pars{\phi} - 2}
\end{align}

$$
{\cal F}=r - \half\,\mu r^{4}\sin^{2}\pars{2\theta}\sin\pars{2\phi} + 8\mu
$$

$$
\begin{array}{rclcrcl}
\partiald{{\cal F}}{r} & = & 0 & \imp & 
1 - 2\mu r^{3}\sin^{2}\pars{2\theta}\sin\pars{2\phi} & = & 0
\\[2mm]
\partiald{{\cal F}}{\theta} & = & 0 & \imp & 
-\mu r^{4}\sin\pars{4\theta}\sin\pars{2\phi} & = & 0
\\[2mm]
\partiald{{\cal F}}{\phi} & = & 0 & \imp & 
-\mu r^{4}\sin^{2}\pars{2\theta}\cos\pars{2\phi} & = & 0
\end{array}
$$

$\ds{\theta \not\in\braces{0,\pi}}$ and $\ds{\phi \not\in\braces{0,\pi,2\pi}}$. That leads to $\ds{\theta =\phi = {\pi \over 4}}$:
$$
2=xyz^{2}={1 \over 8}\,r^{4}\sin^{2}\pars{\pi \over 2}\sin\pars{\pi \over 2}
={r^{4} \over 8}\ \imp\
\begin{array}{|c|}\hline\\
\quad\color{#66f}{\Large r = 2}\quad
\\ \\ \hline
\end{array}
$$
A: Here is another AM-GM based approach:
$\frac{x^2 + y^2 + \frac{z^2}{2} + \frac{z^2}{2}}{4} \ge \sqrt[4]{ \frac{x^2y^2z^4}{4}} = 1$
So the square of the distance is at least $4$, with equality iff $x^2 = y^2 = \frac{z^2}{2} = 1$. Taking signs into account gives you the correct solution set.
