<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] If we take $x=\rho\sin\theta,y=\rho\cos\theta\sin\phi,z=\rho\cos\theta\cos\phi$, then we have to find the minimum of $\rho$ under the constraint:
$$ \rho^3(\sin^3\theta+(\sin^3\phi+\cos^3\phi)\cos^3\theta)=1 $$
or, by switching to the dual problem, the maximum of:
$$ f(\theta,\rho) = \sin^3\theta+(\sin^3\phi+\cos^3\phi)\cos^3\theta.$$
The stationary points of $f(\theta,\rho)$ are given by:
$$ \left\{\begin{array}{rcl}\frac{\partial}{\partial \theta}\,f(\theta,\phi)=-3\cos\theta\sin\theta\left(-\sin\theta+\cos^3\phi\cos\theta+\cos\phi\cos^3\theta\right)&=&0\\\frac{\partial}{\partial \phi}\,f(\theta,\phi)=-3\cos\phi\sin\phi\left(-\sin\phi+\cos^3\phi\cos\theta+\cos\phi\cos^3\theta\right)&=&0\end{array}\right.$$
so, at least in principle, we can locate them. Anyway, since:
$$ 1-\left(\cos^3 x+\sin^3 x\right) = 4\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)\sin^2\left(\frac{x}{2}\right)\left(2+\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)\right)\geq 0$$
$$ 1+\left(\cos^3 x+\sin^3 x\right) = 4\sin^2\left(\frac{\pi}{4}+\frac{x}{2}\right)\cos^2\left(\frac{x}{2}\right)\left(2-\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)\right)\geq 0$$
the range of $g(x)=\sin^3 x+\cos^3 x$ is $[-1,1]$, so the maximum of $f(\theta,\phi)$ is one and the minimum of $\rho$ is $\color{red}{1}$, too. Obviously, it is attained at $(1,0,0),(0,1,0),(0,0,1)$.
For non-believers, here it is a picture of the unit sphere with our surface being a nice hat:

</s>