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 $m$.
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.