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Math Help - Minimize

  1. #1
    Senior Member Sampras's Avatar
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    Minimize

    Suppose you want to minimize  a_{1}^{2} + \cdots + a_{n}^{n} subject to the constraints  a_i >0 and  a_1+ \cdots + a_n = 1 .

    So in the Cauchy Schwarz Inequality do the following: Take  b_1 = \cdots = b_n = 1 . So we have:

     (a_{1}^{2}+ \cdots + a_{n}^{2})(b_{1}^{2} + \cdots b_{n}^{2}) \geq (a_{1}b_{1} + \cdots +a_{n}b_{n})^2 .

     a_{1}^{2} + \cdots + a_{n}^{2} \geq (a_1+ \cdots + a_n)^2 .

     a_{1}^{2} + \cdots + a_{n}^2 \geq 1


    So  a_1 = \cdots = a_n = \frac{1}{\sqrt{n}} ?
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  2. #2
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Sampras View Post
    Suppose you want to minimize  a_{1}^{2} + \cdots + a_{n}^{n} subject to the constraints  a_i >0 and  a_1+ \cdots + a_n = 1 .

    So in the Cauchy Schwarz Inequality do the following: Take  b_1 = \cdots = b_n = 1 . So we have:

     (a_{1}^{2}+ \cdots + a_{n}^{2})(b_{1}^{2} + \cdots b_{n}^{2}) \geq (a_{1}b_{1} + \cdots +a_{n}b_{n})^2 .

     a_{1}^{2} + \cdots + a_{n}^{2} \geq (a_1+ \cdots + a_n)^2 .

     a_{1}^{2} + \cdots + a_{n}^2 \geq 1


    So  a_1 = \cdots = a_n = \frac{1}{\sqrt{n}} ?
    Nevermind I forgot that  b_1^{2} + \cdots + b_{n}^{2} = n .
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  3. #3
    Senior Member Shanks's Avatar
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    Yes, you've got closer to the correct solusion, actually a_1=a_2=\cdot \cdot \cdot =a_n=\frac{1}{n}.
    P.S. you are so humorous.
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