Use Lagrange multipliers to find the point on the surface
$$\frac1x + \frac1y + \frac1z =1$$ which is closest to the origin.
I was wondering if I would start off by using the distance formula, $$d=\sqrt{\left(\frac1x\right)^2+\left(\frac1y\right)^2+\left(\frac1z\right)^2}$$ which would then simplify to, $$d^2=\left(\frac1x\right)^2+\left(\frac1y\right)^2+\left(\frac1z\right)^2$$
I don't know where to go from there, so if someone could help me out, that would be great.
Best Answer
I don't have time to work out all the details in an answer, but here's a quick starter.
The key idea behind Lagrange multipliers is that when two surfaces are tangent to each other, their normal vectors at that point are parallel. In this case, you want to find when the surface $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ is closest to the origin, that is, when it is tangent to a sphere of some radius.
Note first that if $F(x,y,z) = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$, the surface is equal to $F^{-1}(1)$, so the normal field is given by the gradient $\nabla F$.
It's easier to use the energy function $d^2(x,y,z) = x^2 + y^2 + z^2$, because square roots are silly. To find where distance from the origin is minimized, find the gradient of the distance function, $\nabla (d^2)$, and find where it points the same direction as $\nabla F$: $$\nabla(d^2) = \lambda \nabla F,$$ subject to $F = 1$.