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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?

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    $\begingroup$ Do you mean the minimum value? Because $(x,y,z) = (n, 1/n,1)$ can give you arbitrarily large sums $\endgroup$ – Krishnarjun Nov 28 '20 at 14:48
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    $\begingroup$ 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. $\endgroup$ – saulspatz Nov 28 '20 at 14:48
  • $\begingroup$ For finding the min, you can use the Lagrange multiplier ! $\endgroup$ – Anthony Saint-Criq Nov 28 '20 at 14:54
  • $\begingroup$ Sorry for the critical typo. I did mean minimum. $\endgroup$ – Cyh1368 Nov 28 '20 at 15:00
  • $\begingroup$ Does this answer your question? Minimize $P(x,y,z)=(2x+3y)(x+3z)(y+2z)$, when $xyz=1$ $\endgroup$ – Anthony Saint-Criq Nov 28 '20 at 15:15
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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.

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  • $\begingroup$ Furthermore, is there a way to find the values for $x, y$ and $z$ when $x+y+z=3$? $\endgroup$ – Cyh1368 Nov 28 '20 at 15:21
  • $\begingroup$ @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 $0$. $\endgroup$ – Jethalal Nov 28 '20 at 15:39
  • $\begingroup$ 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. $\endgroup$ – Macavity Nov 28 '20 at 16:03
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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.

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