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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??2012-04-13
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    sorry, that's the topic of another question,, i forgot to change the title2012-04-13
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    Did you try to do the cases $n=2$ and $n=3$? They should give a good indication.2012-04-13
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    well, may be you get it wrong, it is not an induction question,2012-04-13
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    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.2012-04-13
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    o, i see. Indeed, i have tried show that the domain of f is unbounded and hence it doesn't have global max but don't know whether it is right. Also for global min, actually i claim that f(x)=1 is the global min and use contradiction to prove 1 must be the global minimum, is it correct?2012-04-13
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    Yes, this is correct and the global minimum $1$ is assumed exactly at $x=(1,\ldots,1)$. The point is that the level sets of $f$ are hyperplanes perpendicular to this vector and that the constraint $1 = g(x_1,\ldots,x_n) = x_1\cdots x_n$ describes a hyperboloid which is tangent to the plane $f(x) = 1$ exactly at $(1,\ldots,1)$. Try to see it in two and three dimensions. To make the calculation rigorous, you *can* use Lagrange multipliers, as suggested in Riemann's answer, which would be the standard brute-force attack on this problem.2012-04-13
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    if it exists a global maximum or global min it must be able to tell by looking at its properties without using lagrange multipliers although we may not tell which point the the global extremum points.2012-04-13
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    That should have been global minimum $n$, of course. Sorry about that. The inequality Riemann mentions in the answer is called the [AM-GM inequality](http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means) and it has many proofs that avoid Lagrange multipliers.2012-04-13

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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 domain2012-04-13
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    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.2012-04-13
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    There is one exception to this, if n=1 then there is a global maximum.2012-04-13
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    yes, of course! But $n=1$ is trivial.2012-04-14
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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.2012-04-15
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    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.2012-04-15