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Let $n$ be a positive integer and $x:=(x_1,x_2,\ldots,x_n)$. For non-negative $x_1,x_2,\ldots,x_n$, consider the function value of $f(x)=x_1+x_2+\cdots+x_n$ subject to the constraint $x_1x_2\cdots x_n=1.$ What I want to know is, does $f(x)$ have:

  1. A global maximum?

  2. A global minimum?

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What does your title have to do with the body of the question?? – The Chaz 2.0 Apr 13 '12 at 15:09
sorry, that's the topic of another question,, i forgot to change the title – Mathematics Apr 13 '12 at 15:11
Did you try to do the cases $n=2$ and $n=3$? They should give a good indication. – t.b. Apr 13 '12 at 15:14
well, may be you get it wrong, it is not an induction question, – Mathematics Apr 13 '12 at 15:16
I wasn't talking about induction at all. But those cases are very tractable and give a good idea of what happens in the general case. – t.b. Apr 13 '12 at 15:17

Maybe you can see that PROBLEM from this inequality:$\frac{x_1+x_2+\cdots+x_n}{n}\geq \sqrt[n]{x_1x_2\cdots x_n}=1$, so $x_1+x_2+\cdots+x_n\geq n$, it has global minimum $1$,but it does not have global maximum, because $x_1+x_2+\cdots+x_n$ can be greater than any given positive number $M$. To see this you can take $x_1=M,x_2=1/M, x_3=1,\ldots,x_n=1.$

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of course i did, however, we should be at least able to conclude whether it has a global max or min by just looking at the domain – Mathematics Apr 13 '12 at 15:24
Maybe you can see that $\frac{x_1+x_2+……+x_n}{n}\geq \sqrt[n]{x_1x_2……x_n}=1$, so $x_1+x_2+……+x_n\geq n$,it has global minimum and does not have global maximum. – Riemann Apr 13 '12 at 15:44
There is one exception to this, if n=1 then there is a global maximum. – JB King Apr 13 '12 at 16:26
yes, of course! But $n=1$ is trivial. – Riemann Apr 14 '12 at 1:54

It has no global maximum, since, for example, you can let $x_3=x_4=\cdots=x_n$ and $x_2=1/x_1$, and then the product is $1$ and the sum is bigger than $x_1$, and you can make $x_1$ as big as you want.

If you draw the picture in the case $n=2$, you'll probably expect it does have a global minimum. When $x_1=\cdots=x_n=1$, then $x_1+\cdots+x_n=n$, and if you're not sure that's the minimum, just consider the set of all points where the sum is $\le 1$, and try to reason to the conclusion that that's a compact set. The whole graph of $x_1\cdots x_n=1$ is a closed set since it's the inverse image of a single point under a continuous function. Once a set in Euclidean space is closed and bounded, it's compact.

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why the inverse image of a single point under a continuous function closed? As i remember, the complement of open is close. – Mathematics Apr 15 '12 at 1:36
The function $(x_1,\ldots,x_n)\mapsto x_1\cdots x_n$ is continuous. The graph of $x_1\cdots x_n=1$ is the inverse-image, under that function, of the set $\{1\}$, which is closed. Therefore the graph itself is a closed set. – Michael Hardy Apr 15 '12 at 18:14

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