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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!

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up vote 1 down vote accepted

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 $\infty$.

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It really is a good idea in this kind of question to try some things out, as this can give you ideas. Try all the numbers equal. Try some other "small" numbers - does the value of $x+y+z$ increase or decrease when the numbers are different. What happens if you try to make one of the numbers very small? What happens if you make one of them very large (how big can you get)? That is very quick to do, and gives some clues as to which way the inequalities go (easy to get them the wrong way round). – Mark Bennet Nov 2 '12 at 21:48
How do you know that infimum is indeed 3? By subtituting $\frac{1}{n}$ and $n$ for $x, y$ we can go as close to 3 as we want to, but does $x*y*z=1$ directly imply that infimum is 3? Is it possible to prove with definitions, for example with epsilon? – fdhd Nov 2 '12 at 21:51
@user46034 The AM-GM gives us $x+y+z \geq 3$. Equality holds at $x=y=z = 1$. Hence $3$ is indeed the infimum. And I set $x = n, y = \dfrac1n$ and $z = 1$ to argue that the supremum is infinity by letting $n \to \infty$. – user17762 Nov 2 '12 at 21:53
O my God... I see it now. Thank you a lot – fdhd Nov 2 '12 at 21:56

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