Supremum, infimum I came across nasty task which includes supremum, infimum and I am confused about it. The question is to find a supremum and infimum of a given set:$$A =(x+y+z:x,y,z>0,xyz=1)$$I tried to eliminate z and use corelations between arithmetic mean and so on. I get $A =(\frac{xy(x+y)+1}{xy}:x,y>0)$, and I know that $\frac{x+y}{2}>\sqrt{xy}>\frac{2}{\frac{1}{x}+\frac{1}{y}}=\frac{2xy}{x+y}$. From this I get $xy<\frac{1}{4}(x+y)^{2}$ $(x+y)^2>2\sqrt{xy}$, what after substitution gives $$\frac{2(xy)^{\frac{3}{2}}+1}{xy}<\frac{xy(x+y) +1}{xy}<\frac{(x+y)^3 +4}{(x+y)^2}$$but I have no idea how to evaluate it (if it is a good way). Intuition says that infimum is $0$, and supremum $\infty$, but how to prove it more formally? Thanks in advance!

For infimum, recall AM-GM. Given $x,y,z > 0$, we have that
$$\dfrac{x+y+z}3 \geq \sqrt[3]{xyz}$$
and equality holds when $x=y=z$. Since $xyz = 1$, we get that $x+y+z \geq 3$. Hence, the infimum is $3$.
For supremum, consider $x = n, y = \dfrac1n$ and $z = 1$, where $n$ can be arbitrarily large. Note that $xyz = 1$. $x+y+z = n + \dfrac1n + 1$. Hence, the supremum is $\infty$.
