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The Arithmetic-Geometric Mean Inequality Problem

  1. Oct 31, 2009 #1
    1. The problem statement, all variables and given/known data
    verify that f(x) = log(1+ex) is convex and use this to show that
    (1+x1)1/n(1+x2)1/n..(1+xn)1/n >= 1 + (x1x2...xn)1/n
    where x1,x2,...xn are positive real numbers.


    2. Relevant equations



    3. The attempt at a solution
    I know f(x) is convex b/c f''(x) is always positive.
    and I can guess the above ineqality is true, but
    I do not konw how I use the f(x) to prove the inequality.
    Can log (1+e^x) imply to (1+x1)1/n...(1+xn)1/n..?
     
  2. jcsd
  3. Oct 31, 2009 #2

    Dick

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    Start by writing xi=e^(log(xi) and try to express both sides of the inequality in terms of the function f(x). Taking the log of both sides would be a good start.
     
  4. Oct 31, 2009 #3
    Thanks,
    so.. f(x) = (1+ex)

    (1+x1)1/n...(1+xn)1/n =
    (1+elnxn)1/n ==> f(x)
    Thus,
    Σ((1/n)log(1+elnxi)) = Sigma (f(x)*(1/n))

    Also,
    log (1+ (ex1...xn)1/n
    = log (1 + eln Σxi(1/n))
    Therefore,

    Σ(f(x)*(1/n)) >= f(Σ(1/n)*xi) as required. right?

    I am not sure
    log (1 + eln Σ(1/n)*xi) implies to f(Σ(1/n)*xi).
     
  5. Oct 31, 2009 #4

    Dick

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    I'm not sure that is quite correct, since you seem to have dropped the log's at the end. But one side should be the mean of f(log(xi)) and other side should be f(mean log(xi)) where 'mean' is the arithmetic mean. If f is convex, what's the relation between those two? Hint: if f is convex then f((a+b)/2)<=(f(a)+f(b))/2. That can be generalized to more than just two values a and b.
     
    Last edited: Oct 31, 2009
  6. Nov 1, 2009 #5
    Sorry, my writing was confused.

    I konw Jensen’s inequality.
    Suppose that f(x) is a convex function defined on a convex subset C of Rn.
    If a1,...,ak are nonnegative numbers with sum 1 and if x(1),...,x(k) are points of C, then
    f(∑aix(i)) <= ∑aif((i)).

    f(x) = log(1+ex)
    (1+x1)1/n...(1+xn)1/n >= 1 + (x1..xn)1/n

    (1+x1)1/n...(1+xn)1/n -->
    (1/n)log(1+elog(x1))+...(1/n)log(1+elog(xn)) = ∑(1/n)log(1+elog(xi))

    1 + (x1..xn)1/n -->
    log(1 + elog(x1..xn)1/n) = log(1 + e(1/n)log(x1)+..(1/n)log(xn)) = log(1 + e∑(1/n)log(xi))

    Therefore,
    ∑(1/n)log(1+elog(xi))>= log(1 + e∑(1/n)log(xi))
    This inequality is equivalent to the Theorem as required inequality.

    Right?
     
  7. Nov 1, 2009 #6

    Dick

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    That's it.
     
  8. Nov 1, 2009 #7
    Thanks!!
     
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