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An inequality to prove

  1. Nov 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Let [itex] 0 ≤ x_1 ≤ x_2 ≤ ... ≤ x_n [/itex] and [itex] x_1 + x_2 + ... + x_n = 1 [/itex] , All the 'x' are real and n is a natural number. Prove the following:

    [tex] (1+x_1^21^2)(1+x_2^22^2)...(1 + x_n^2n^2) ≥ \frac{2n^2+9n+1}{6n} [/tex]


    2. Relevant equations


    3. The attempt at a solution

    Well....all I managed to do was to "check the field a little". First of all, the right member of the inequality somehow reminds me of the sum:

    [tex] 1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n^2 + 3n + 1)}{6} [/tex]

    I know from experience that we can solve this type of exercises by using some well-know inequalities such as the one that relates the average mean with the geometric one, or the Cauchy-Buniakowski-Schwarz ( C-B-S ). So:

    [tex]1 ≥\frac{1}{n} ≥ \sqrt[n]{x_1x_2...x_n} [/tex] , so it follows immediately that : [tex] x_1x_2...x_n ≤ 1[/tex]

    That's all I could do. Could you help me continue, please? This exercise has been stuck in my head for two weeks already. I can't solve it alone, and I would like that somebody could guide me to the solution. Thank you!
     
    Last edited: Nov 26, 2013
  2. jcsd
  3. Nov 26, 2013 #2

    Ray Vickson

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    Are all the ##x_i \geq 0?##
     
  4. Nov 26, 2013 #3
    Yes, they are. I also modified the first post, by adding this fact.
     
  5. Nov 27, 2013 #4
    Can you help me ? Do you have any idea :D ?
     
  6. Nov 27, 2013 #5

    haruspex

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    This is under precalculus. If calculus methods are permitted I would use Lagrange multipliers.
     
  7. Nov 27, 2013 #6

    PAllen

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    You can make elementary (non calculus) arguments that the left side of the inequality is ≥ (1+n). Then, you can get the rest by algebra.
     
  8. Nov 27, 2013 #7

    Ray Vickson

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    One can solve the optimization problem
    [tex] \min \sum_{i-1}^n \log(1 + i^2 x_i^2) \\
    \text{subject to}\\
    x_1 + x_2 + \cdots + x_n = 1 \\
    0 \leq x_1 \leq x_2 \leq \cdots \leq x_n[/tex]
    using a numerical optimization package for various small to medium values of n (3 ≤ n ≤ 10). In all cases I tried the optimal solution is ##x_1 = x_2 = \cdots = x_n = 1/n##. Furthermore, this can be shown to be a so-called Kuhn-Tucker point for the problem with general n. The minimization objective is pseudo-convex and the constraint set is convex, so satisfying the Karush-Kuhn-Tucker conditions is, I believe, sufficient for a (global) minimum.

    In other words, we have
    [tex] (1+1^2 x_1^2)(1 + 2^2 x_2^2) \cdots (1+ n^2 x_n^2)
    \geq (1 + (1/n)^2)(1+(2/n)^2) \cdots (1 + (n/n)^2)[/tex]
    for all feasible x. Maybe that can be manipulated further to give the desired result.
     
  9. Nov 27, 2013 #8

    haruspex

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    This is confusing. 'n' is the number of terms, so the nth term means just the last term. Certainly xn >= 1/n, so the last term is at least 2. I don't see how you get it's at least 1+n.
    If you meant a more generic n, I still don't see it. xj need not be ≥ 1/j, nor 1/n. Every xj except xn could be zero.
     
  10. Nov 27, 2013 #9

    PAllen

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    Sorry, I realized my mistake as soon as finished, and was working on deleting my post. I missed that the x's were squared as well as the i's. Oops.
     
  11. Nov 27, 2013 #10

    haruspex

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    Suppose xj < xj+1 for some j < n. It might be straightforward to show that substituting xj' = xj+1' = (xj + xj+1)/2 produces a lower product. (Calculus by the back door.)
     
  12. Nov 27, 2013 #11

    haruspex

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    Yeah, I've been there - a frantic dash to delete a regretted post, only to find it's already been read:redface:.
     
  13. Nov 27, 2013 #12

    PAllen

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    Unfortunately, this is not true. The substitution produces a larger product. Larger by the difference between xj and its successor, over 2, squared.
     
  14. Nov 27, 2013 #13

    haruspex

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    Hmm.. I just tried it and it worked for me. I ended up needing that 1 > j(j+1)xjxj+1, which is clearly true.
     
  15. Nov 27, 2013 #14

    Ray Vickson

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    A much easier way to get a lower bound is to note that for positive ##y_i = i^2 x_i^2## we have
    [tex] (1+y_1)(1+y_2) \cdots (1+y_n) > 1 + \sum_{i=1}^n y_i, [/tex]
    because the RHS is just the first two terms in the expansion of the product, and all the omitted terms are > 0. Therefore, we can get a lower bound by solving the relatively simple problem
    [tex] \min \sum_{i=1}^n i^2 x_i \:\:\:\leftarrow \sum_{i=1}^n y_i\\
    \text{subject to}\\
    \sum_{i=1}^n x_i = 1\\
    0 \leq x_1 \leq x_2 \leq \cdots \leq x_n[/tex]
    This problem has the optimal solution ##x_1 = x_2 = \cdots = x_n = 1/n##, so we have
    [tex] (1+1^2 x_1^2)(1+2^2 x_2^2) \cdots (1 + n^2 x_n^2) > 1 + \sum_{i=1}^n i^2/n^2 [/tex]
    for all feasible ##x_1, x_2, \ldots x_n##.
     
    Last edited: Nov 27, 2013
  16. Nov 27, 2013 #15

    PAllen

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    I keep forgetting the squaring.
     
    Last edited: Nov 27, 2013
  17. Nov 27, 2013 #16

    PAllen

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    This seems promising. This lower bound appears to equal the RHS. If this could be proven (along with the rest), you'd be done.
     
  18. Nov 28, 2013 #17
    We need the proof for this in elementary terms.
    If this is proven, the problem is basically solved.

    Edit: I just saw that the [itex]x_i[/itex] had to be positive. Then, the proof is obviously very easy.
    The sum [itex]\displaystyle \sum_{i=1}^n i^2/n^2[/itex] has closed form [itex]\displaystyle \frac{(n+1)(2n+1)}{6n}[/itex]. The inequality then follows from the optimal solution.
     
    Last edited: Nov 28, 2013
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