<s>[INST] Points on Surface, Distance Optimized How do I find the points on the surface:
$$x^3+y^3+z^3=1$$
such that the distance to the origin is minimized?
My Thoughts:
Perhaps we can minimize the distance squared? Not sure.
 [/INST] You want to minimise the function: $f(x,y,z) = x^2+y^2+z^2$ with the condition that $g(x,y,z) = 1- x^3-y^3-z^3 = 0$
Note that if $x^2+y^2+z^2$ is minimal so is $\sqrt{x^2+y^2+z^2}$
This is best done using Lagrange multipliers. You define $ F(x,y,z,\lambda ): = f(x,y,z)  + \lambda\cdot  g(x,y,z)$
Then the set of equations 
$$ \frac{\partial F}{\partial x} = 0,\ \frac{\partial F}{\partial y} = 0,\ \frac{\partial F}{\partial z} = 0,\ \frac{\partial F}{\partial \lambda}= 0
$$
will give us the answer. We get $x=\lambda 3x^2$ same for $y,z$ plus $g(x,y,z) = 0$
(i) $x,y,z \ne 0$
We get: $2x=\lambda 3x^2 \Rightarrow \lambda = \frac{2}{3x}$ 
Inserting into the second and third equation: $x = y = z$
With the last equation: $3x^3 = 1 \Rightarrow x =y=z = \frac{1}{\sqrt[3]{3}}$
(ii) $x,y\ne 0$ and $z = 0$
$ 2x^3 = 1 \Rightarrow x =y = \frac{1}{\sqrt[3]{2}}, z = 0$
(iii) $x\ne 0, y,z = 0$
$ x^3 = 1 \Rightarrow x = 1, y = z = 0$ and $\lambda = \frac23$.
Because of symmetry we will get the last two cases three times for the permutations of $x,y,z$
Now we calculate the distances:
(i) $d = 3^{\frac{1}{6}} = 1,2...$
(ii) $d = 2^{\frac{1}{6}} = 1,1...$
(iii) $d = 1$
So the points with minimal distance are case (1) $ x = 1, y=z=0$ (2) $ y = 1, x=z=0$ (3) $ z = 1, x =y=0$.
To check if we have a local minimum we could also calculate the Hessian matrix on the tangent space of $\nabla g(x,y,z) = (-3, 0, 0)^T$ for say case (1):
\begin{align}
\frac{\partial^2F}{\partial y^2} = \frac{\partial^2F}{\partial z^2} &= 2 - 6\lambda y =  2 \\
\frac{\partial^2F}{\partial x \partial y} &= 0
\end{align}
That is
$$ \nabla^2_{(y,z)} f(1,0,0) = \begin{pmatrix} 2 & 0 \\ 0 & 2  \end{pmatrix}.$$
This is obviously positive define, thus $x=1,y=z=0$ is a strict local minimum.
Here is an alternative solution for the minima argument, not sure if this is correct though:
If I vary $y,z$ to be slightly different from zero: $y = \epsilon$, $z = \delta$ then x is given by $x = (1-\epsilon^3 -\delta^3)^{\frac{1}{3}}$. Now checking the distance: $ d= (1-\epsilon^3 -\delta^3)^{\frac{2}{3}} + \epsilon^2 +\delta^2$ tayor expanding the first term $d= 1 - \frac{2}{3}\epsilon^3 - \frac{2}{3}\delta^3 + \epsilon^2 +\delta^2 =  1 + \epsilon^2(1- \frac{2\epsilon}{3}) +\delta^2(1- \frac{2\delta}{3}) > 1$
similarly for just varying one component. Thus the points are indeed local minima. 
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