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Math Help - Interesting problem, any proofs?

  1. #1
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    Interesting problem, any proofs?

    Let x_{1},x_{2},...,x_{n} be positive real numbers. Prove that

    n^2\leq(x_1+x_2+...+x_n)(\frac{1}{x_1}+\frac{1}{x_  2}+...+\frac{1}{x_n}) and

    \frac{x_1+x_2+...+x_n}{\sqrt{n}}\leq\sqrt{(x_1)^2+  (x_2)^2+...+(x_n)^2}

    I really have no idea how to even begin this... any ideas?
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  2. #2
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    Quote Originally Posted by paupsers View Post
    Let x_{1},x_{2},...,x_{n} be positive real numbers. Prove that

    n^2\leq(x_1+x_2+...+x_n)(\frac{1}{x_1}+\frac{1}{x_  2}+...+\frac{1}{x_n}) and

    \frac{x_1+x_2+...+x_n}{\sqrt{n}}\leq\sqrt{(x_1)^2+  (x_2)^2+...+(x_n)^2}

    I really have no idea how to even begin this... any ideas?
    They are both consequence of Cauchy-Schwarz inequality: |\vec{x}\cdot\vec{y}|\leq \|\vec{x}\|\|\vec{y}\|. I let you find what to take as \vec{x} and \vec{y}.
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  3. #3
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    I've been looking at this problem for a while, and I can't figure out how to isolate the n's on one side, and how that is going to relate to the right-hand side.

    Any additional help?
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  4. #4
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    Quote Originally Posted by paupsers View Post
    I've been looking at this problem for a while, and I can't figure out how to isolate the n's on one side, and how that is going to relate to the right-hand side.

    Any additional help?
    The Cauchy-Schwarz inequality expressed for sums is:

    \left(\sum_{i=1}^n a_ib_i\right)^2\leq\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right)

    Try letting a_i=\sqrt{x_i} and b_i=\frac{1}{\sqrt{x_i}}

    You can apply similar reasoning to the second problem.
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  5. #5
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    I understand the first problem now, but I'm having a really hard time finding a_i and b_i for the second one. Can you explain the method of how to find them for the second one so I can use that method for the rest of the problems in my book?
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  6. #6
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by paupsers View Post
    I understand the first problem now, but I'm having a really hard time finding a_i and b_i for the second one. Can you explain the method of how to find them for the second one so I can use that method for the rest of the problems in my book?
    Take the square root of the C-S Inequality:

    \sum_{i=1}^n a_ib_i\leq\sqrt{\sum_{i=1}^n a_i^2}\sqrt{\sum_{i=1}^n b_i^2}

    Now rearrange your problem a bit:

    (x_1+x_2+...+x_n)\leq\sqrt{(x_1)^2+(x_2)^2+...+(x_  n)^2}\sqrt{n}

    Do you see what to do now?
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  7. #7
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    Does a_i=x_1+x_2+...+x_n and b_i=1?
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  8. #8
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by paupsers View Post
    Does a_i=x_1+x_2+...+x_n and b_i=1?
    Yes. a_i=x_i (not x_1+x_2+...) and b_i=1.
    Last edited by redsoxfan325; October 31st 2009 at 09:15 AM.
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    Thanks a lot for the help!
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  10. #10
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    for question one , i attempt to prove this by Mathematical Induction .

    When  n=1

    1^2 =  \frac{1}{x_1} \cdot  ( x_{1} )



    Let  P = x_1 + x_2 + ... + x_k  , Q = \frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_{k}}

    We assume  PQ \geq k^2

    When  n = k+1

     ( x_1 + x_2 + ... + x_k + x_{k+1} ) \cdot ( \frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_{k}} + \frac{ 1}{x_{k+1}} ) = (P + x_{k+1} ) (Q + \frac{1}{x_{k+1}}) =  PQ + 1 + \frac{P}{x_{k+1}} + x_{k+1} Q \geq k^2 + 1 + 2\sqrt{PQ} \geq k^2 + 2k + 1 = (k+1)^2
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