Finding maximum of $x+y+z$ [closed]

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?

• Do you mean the minimum value? Because $(x,y,z) = (n, 1/n,1)$ can give you arbitrarily large sums – Krishnarjun Nov 28 '20 at 14:48
• There is no maximum. Let $z=1$, y=$\frac1x$. Then $x+y+z=x+1+\frac1x$ and we can take $x$ as large as we please. – saulspatz Nov 28 '20 at 14:48
• For finding the min, you can use the Lagrange multiplier ! – Anthony Saint-Criq Nov 28 '20 at 14:54
• Sorry for the critical typo. I did mean minimum. – Cyh1368 Nov 28 '20 at 15:00
• Does this answer your question? Minimize $P(x,y,z)=(2x+3y)(x+3z)(y+2z)$, when $xyz=1$ – Anthony Saint-Criq Nov 28 '20 at 15:15

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.

• Furthermore, is there a way to find the values for $x, y$ and $z$ when $x+y+z=3$? – Cyh1368 Nov 28 '20 at 15:21
• @Cyh1368 There's no need to find the values of x, y, and z. Further, there is only $1$ possible value $(1,1,1)$ of $x,y,z$ satisfying $x+y+z=3$ for $x,y,z>0.$. Note $x,y,z>0.$ All x,y,z are positive, that is greater than $0$. – Jethalal Nov 28 '20 at 15:39
• From positivity condition, we could just say $x+y+z\ge 0$ and conclude that's the minimum. Of course no values satisfying all other conditions will reach that minimum, hence it's just a lower bound. The difference between a lower bound and a minimum is that the latter can actually be achieved. Hence finding at least one set of $x,y,z$ that gets equality is important. – Macavity Nov 28 '20 at 16:03

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.