Minimising distance from origin to surface

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I am trying to minimise the distance from the origin to the surface $3x^2+\sqrt{3}y^2-\sqrt{3}z^2=1$.

I tried using this constraint and writing $z^2=\sqrt{3}x^2+y^2-\frac{1}{\sqrt{3}}$, and then substituting this into $\sqrt{x^2+y^2+z^2}$.

This gives a new function to minimise, $f(x,y)=\sqrt{(1+\sqrt{3})x^2+2y^2-\frac{1}{\sqrt{3}}}$.

But this function can equal zero for some values of $x$ and $y$ (I checked on Desmos too) and I know that no points on the surface have a distance from the origin equal to zero. Where have I gone wrong?

I've looked at similar questions and it looks like there's another way of answering questions like this using Lagrange multipliers, but I am not familiar with those and I think there should be a way to make this method work, but I can't figure out what mistake I've made.

Thanks for your help!

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Rewrite the constraint as $$x^2 = \left( 1 - \sqrt{3}y^2 + \sqrt{3} z^3\right) / 3$$ You want to find a point on the curve with minimal norm. So we are minimizing norm, but we might as well minimize the square of the norm, to get rid of the square root in the norm. So, the problem you want to solve is $$\min_{x,y,z \in \mathbb{R}} x^2 + y^2 + z^2 \quad \text{ s.t. } \quad 3x^2 + \sqrt{3} y^2 -\sqrt{3} z^2 = 1.$$ Now we perform the above substitution of $x^2$ in the objective function. Implicitly, this places a constraint on the values of $y$ and $z$, namely $$\sqrt{3}y^2 - \sqrt{3}z^2 = 1 - 3x^2 \leq 1.$$ Hence, after substitution, the problem you are left with is $$\min \left( 1 - \sqrt{3}y^2 + \sqrt{3} z^3\right) / 3 + y^2 + z^2 \quad \text{ s.t. } \quad \sqrt{3} y^2 - \sqrt{3} z^2 \leq 1.$$ Note that the objective function is convex (verify this yourself), so if we can find a stationairy point (all derivatives equal $0$) that is feasible, we have solved the problem. It turns out (check the details yourself) that the stationairy point $(y,z) = (0,0)$ is feasible, so this is the global minimum! Solving for $x$ gives $x = \pm 1 /\sqrt{3}$, and hence $$(x,y,z) = ( \pm 1/\sqrt{3},0,0)$$ are the two points on the curve with minimal norm.