Finding maximum of $x+y+z$

Question

If positive numbers $x, y$ and $z$ satisfy that $xyz=1$, what is the minmum value for $x+y+z$?

From $xyz=1$, we can get $$x = \frac{1}{yz};\space\space\space y = \frac{1}{xz};\space\space\space z = \frac{1}{xy}; $$

Subsitute them into $x+y+z=1$ and I got$$\frac{xy+yz+xz}{xyz} = xy+yz+xz = 1$$

Since we're finding the minimum for $x+y+z$, I thought of using the formula $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+xz)$ due to the fact that we have the value of $xy+yz+xz$.

That's all I've got so far. How can I continue?

Answer 1

Use AM-GM inequality,

$$\frac{x+y+z}{3} \ge \sqrt [3]{xyz}$$

$$x+y+z \ge 3$$

The minimum is $3$ and there's no maximum.

Answer 2

By geometry:

The surface of equation $xyz=1$ (don't know its name) is a cubic with an "hyperbolic-like" shape, as any cross section by a plane of one constant coordinate is an hyperbola. It has a symmetry of order $3$ around the axis $x=y=z$, and is open towards infinity.

The sections by the plane $x+y+z=c$ are closed curves, starting from $c=3$ and enlarging monotonously and unboundedly.

The minimum is $c=3$ and there is no maximum.