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 $F = 1$.
let $P(rcost,rsint)$ be a Point on Hyperbola. so its distance from $(0,0)$ is $r$, so we need to find Minimum value of $r$. Since $P$ lies on Hyperbola
$$r^2cos^2t+8r^2sintcost+7r^2sin^2t=225$$ $\implies$
$$r^2=\frac{450}{8sin2t-6cos2t+8}$$
Now max value of $$8sin2t-6cos2t+8$$ is $$\sqrt{8^2+6^2}+8=18$$
hence Min value of $r^2$ is $\frac{450}{18}=25$
So shortest distance is $5$
Best Answer
I do not think you need LaGrange multiplier. Define $f(x)$ to be the square of the distance between the point $(2,0)$ and the hyperbola. (It is enough to minimize the square of the distance to make the calculus easier). Using the top half of the hyperbola, with a bit of math we get $$f(x)=(x-2)^2+(\sqrt{x^2+4}-0)^2.$$ Minimize this function using the first derivative test to find the value of $x$ on the hyperbola closest to $(2,0).$ Then use the hyperbola equation to find the $y-$value.