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.
Best Answer
AM-GM:
$x,y >0;$
$d^2=z^2+x^2+y^2=$
$(1/xy)^2+ x^2+y^2 \ge 3(1)^{1/3}=3$.
Attained for $x^2=y^2=(1/xy)^2$, for $x=y=1$.