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If the product of n numbers is 1, their sum is >= n

  1. Jun 29, 2007 #1
    i'm reading shilov's "real and complex analysis". there is a problem to prove that for real positive x1,x2,...xn, if x1*x2*...*xn=1, then x1+x2+...xn >= n. I proved this for the case n = 2. it says use induction on this case to prove it for the nth case. but i just don't see it.

    for n=2:

    (a+1)^2 >= 0
    a^2 + 2a + 1 >= 0
    a^2 + 1 >= 2a
    a + 1/a >= 2
  2. jcsd
  3. Jun 29, 2007 #2
    Use the AM-GM inequality.

    [tex]\frac{x_1+x_2+...+x_n}{n} \geq \sqrt[n]{x_1x_2...x_n} = 1[/tex]

    [tex]x_1+...+x_n \geq n[/tex]
  4. Jun 29, 2007 #3
    right. actually it's a later problem to prove that inequality. but that's not really what i'm looking for. he asks as one problem to prove that a + 1/a >= 2. then he says use induction to prove it for the general case. so what i really want is an inductive proof.
  5. Jun 30, 2007 #4
    hey, this AM-GM equality gives me a problem, let's suppose we would like to generalize it to an infinite dimensional space, the AM is:

    [tex] \int_{a}^{b}dx (b-a)^{-1} f(x) [/tex]

    however what would be the analogue of the GM ?
  6. Jun 30, 2007 #5
    We need the restriction
    [tex]f(x) >0[/tex]

    Then, perhaps this is a reasonable generalization,
    [tex]\exp \left(\int_a^b \ln f(x) dx\right)[/tex]
  7. Jun 30, 2007 #6


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    That's exactly right, and both inequalities can be proven essentially the same way using http://en.wikipedia.org/wiki/Jensen%27s_inequality" [Broken], as shown in chapter 3 of Rudin's analysis book.
    Last edited by a moderator: May 2, 2017
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