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I have $A = \{x + y + z: x, y, z > 0, xyz = 1 \}$ and I'm investigating whether this set has an infimum and a supremum. It looks to me like there is no supremum as the set doesn't seem to be bounded from above. A candidate for an infimum seems to be $1$. Any hints how I might use $xyz = 1$?

Edit: $${x + y + z \over 3} \ge \sqrt[3]{xyz}$$ $$x+y+z \ge 3$$ $A$ is bounded from below by $3$. Using the definition of infimum, we can prove that $inf \text{A} = 3$.

To prove there is no supremum, we need to show: $$\forall{m>0} \ \exists{x \in A: x>m}$$ This is: $$x+y+z>m$$ If we choose a specific $m$, we can always find three greater natural numbers. Their sum will be greater than $m$.

Zelazny
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    Are you aware of the AM-GM inequality? It will tell you the infimum almost immediately. Also note that you need to demonstrate concretely (or otherwise) that there is no supremum, not just say "it looks like it is unbounded to me". – Arthur Oct 27 '16 at 20:23
  • @Arthur thanks for the hint. I have included this in my original post. – Zelazny Oct 27 '16 at 20:51

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Hint

$$x=m, y=\frac{1}{m}, z=1 \Rightarrow xyz=1$$

Note For the inf, setting $x=y=z=1$ shows that $3$ is a minimum for your set.

N. S.
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  • Technically, setting $x=y=z=1$ only proves that the minimum is at most $3$. Showing that this is the actual minimum requires some argument (and appealing to symmetry alone isn't enough). – Arthur Oct 27 '16 at 21:13
  • @Arthur As the poster said $x+y+z \geq 3$ by the AM-Gm inequality. The observation that this minumum is achieved makes it much easier to prove that it is the inf, instead of "Using the definition of infimum, we can prove that $inf(A)=3$ " – N. S. Oct 27 '16 at 21:31
  • Yes, the AM-GM inequality is sufficient argument. I hadn't noticed it was added to the question. – Arthur Oct 27 '16 at 21:38
  • @Arthur I am always trying to train my students to check first if the set has a max/min before going to the definition of sup/inf.... As for the AM-Gm, I think it was your suggestion, and it was already part of the post when I posted my answer, looking at the timing I think it was added just seconds before I checked it :) – N. S. Oct 27 '16 at 21:42