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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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