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Thread: Minimize

  1. #1
    Senior Member Sampras's Avatar
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    Minimize

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

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

    $\displaystyle (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 $.

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

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


    So $\displaystyle 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 $\displaystyle a_{1}^{2} + \cdots + a_{n}^{n} $ subject to the constraints $\displaystyle a_i >0 $ and $\displaystyle a_1+ \cdots + a_n = 1 $.

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

    $\displaystyle (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 $.

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

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


    So $\displaystyle a_1 = \cdots = a_n = \frac{1}{\sqrt{n}} $?
    Nevermind I forgot that $\displaystyle 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 $\displaystyle a_1=a_2=\cdot \cdot \cdot =a_n=\frac{1}{n}$.
    P.S. you are so humorous.
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