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Math Help - summation question

  1. #1
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    summation question

    i want to show

    1/(n^2) <= (a1^2) + (a2^2) + (a3^2) + ..... + (an^2)

    given that a1 + a2 + .... + an =1

    any idears

    thankss
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  2. #2
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    Quote Originally Posted by silk23 View Post
    i want to show

    1/(n^2) <= (a1^2) + (a2^2) + (a3^2) + ..... + (an^2)

    given that a1 + a2 + .... + an =1

    any idears

    thankss
    Edit: Sorry I misread the problem. I'll think about it. (But someone else will probably answer before I come up with anything.)
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  3. #3
    Senior Member roninpro's Avatar
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    It suffices to show \frac{1}{n}\leq a_1^2+\ldots+a_n^2 since  \frac{1}{n^2}\leq \frac{1}{n}.

    How about using calculus? Lagrange Multipliers might do the trick. Optimize f(a_1,\ldots, a_n)=a_1^2+\ldots+a_2^2 subject to g(a_1,\ldots, a_n)=a_1+\ldots+a_n=1.
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  4. #4
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    ok so let F = (a1^2) + (a2^2) + (a3^2) + ..... + (an^2) + y( a1 + ... + an -1)

    df/dai = 2ai +y = 0

    ai= -y/2

    df/dfy = a1 + ... + an -1 =0

    so y= -2/n so

    ai = 1/n maxamises the (a1^2) + (a2^2) + (a3^2) + ..... + (an^2) so maximum is n*(1/n^2) =1/n

    and 1/(n^2) < 1/n

    thanks
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  5. #5
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    Here is an alternative, non-calculus approach. Assume the contrary, i.e.

    a_1^2 + a_2^2 + \dots + a_n^2 < 1/n^2.

    Then for any i,
    a_i^2 < 1/n^2

    so
    |a_i| < 1/n
    and
    a_i  \leq |a_i|
    so
    a_1 + a_2 + \dots + a_n \leq |a_1| +| a_2| + \dots + |a_n| < 1/n + 1/n + \dots +1/n = 1
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