Point on $z = \frac{1}{xy}$ closest to origin

Question

Where $x>0$ and $y>0$.

I want to work with the square of the distance formula from the origin, so I went with $f(x,y) = x^2 + y^2 + \frac{1}{(xy)^2}$.

Then I found the first partial derivatives:

$f_x = 2x - \frac{2}{x^3y^2}$ and $f_y = 2y - \frac{2}{y^3x^2}$.

Set them equal to $0$ to find the critical points, and multiply both sides by respective $x$ and $y$ to get a common term to set equal to each other:

$2xy = \frac{2}{x^3y}$ and $2xy = \frac{2}{y^3x}$ so $\frac{2}{x^3y} = \frac{2}{y^3x}$. Simplfying, we get $y^2 = x^2$, or $y = x$ since we are given that $x>0$ and $y>0$.

I'm not sure what to do from here.

Answer 1

The surface you are describing can be described as $xyz = 1$. Since this is symmetric in $x$ versus $y$ versus $z$, any point where $|x| = |y| = |z|$ would be an extremum in distance from the origin. Visually, we can see that that such points are minima.

Therefore the solutions must be $x,y,z = (1,1,1), (1,-1,-1), (-1,1,-1),$ or $(-1,-1,1)$. Only one of these has $x,y > 0$.

Answer 2

Alternative solution: The Inequality of arithmetic and geometric means (with $n=3$) states that for non-negative real numbers $a, b, c$

$$ \frac{a + b + c}3 \ge (abc)^{1/3} $$

with equality if and only if $a=b=c$. It follows that if $(x, y, z)$ is a point on the surface $z = 1/(xy)$ then

$$ \frac{x^2 + y^2 + z^2}3 \ge (x^2y^2z^2)^{1/3} = 1 $$

and therefore $x^2 + y^2 + z^2 >= 3$ with equality if and only if $x=y=z$ ($= 1$).