Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]

    So,
    [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

    StatusX

    User Avatar
    Homework Helper

    That's exactly right, and both inequalities can be proven essentially the same way using Jensen's inequality, as shown in chapter 3 of Rudin's analysis book.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: If the product of n numbers is 1, their sum is >= n
  1. N+1 = n(1+1/n) limit (Replies: 3)

  2. Limits, n, n+1 (Replies: 4)

Loading...