I need to find the point on this surface $f(x,y)=\displaystyle \frac{1}{xy}$ where $x,y>0$ that is closest to the origin.
I know intuitively that this would be a normal vector , $a=(x_0,y_0,z_0)$ ,from the surface that passes through the origin. So by working out $\nabla f(x,y)$, I should be able to write an equation: $$\displaystyle \lambda\nabla f(x_0,y_0)=a$$
This is:
$$ \begin{bmatrix} x_0\\ y_0\\ z_0 \end{bmatrix} =\lambda \begin{bmatrix} \frac{-1}{x_0^2y_0}\\ \frac{-1}{x_0y_0^2}\\ 1 \end{bmatrix} $$
But this seems incredibly fiddly and I honestly need to know if there is a better way to do it.
AM-GM is an elegant way to solve this, but I don’t think that solving it via computing the normal to the surface is at all “fiddly.”
Let $F(x,y,z)=f(x,y)-z$. You’re looking for a point on the surface at which the normal is parallel to the vector from the origin to the point. In $\mathbb R^3$ this can be expressed as $(x,y,z)\times\nabla F(x,y,z)=0$. This is equivalent to your Lagrange multiplier formulation, but avoids introducing an additional variable. We compute $$(x,y,z)\times\nabla f(x,y,z) = (x,y,z)\times\left(-\frac1{x^2y},-\frac1{xy^2},-1\right) = \left(\frac z{xy^2}-y,x-\frac z{x^2y},\frac1{x^2}-\frac1{y^2}\right)=0.$$ From the last component and the fact that $x$ and $y$ are both positive, we have $x=y$, and from the equation of the surface, $z=\frac1{xy}$. Making these substitutions reduces this system of equations to the single equation $x^6=1.$