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Math Help - show that the inequality is true

  1. #1
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    show that the inequality is true

    show that ||x||_1 <= \sqrt{n}||x||_2

    I have proved that ||x_2||<=||x_1||

    by definition we have ||x||_1= |x_1|+|x_2|+....|x_n| and ||x||_2=\sqrt{x_1^2+x_2^2+...+x_n^2}

    then we haven ||x_2|| =\sqrt{nx_1^2+nx_2^2+...+nx_n^2} why is this greater than ||x_1||


    IF we square both of them, I get nx_1^2+nx_2^2+...+nx_n^2 and  (|x_1|+|x_2|+....|x_n|)^2 , how to show ||x_2|| si bigger?
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  2. #2
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    Re: show that the inequality is true

    Quote Originally Posted by wopashui View Post
    show that ||x||_1 <= \sqrt{n}||x||_2

    I have proved that ||x_2||<=||x_1||

    by definition we have ||x||_1= |x_1|+|x_2|+....|x_n| and ||x||_2=\sqrt{x_1^2+x_2^2+...+x_n^2}

    then we haven ||x_2|| =\sqrt{nx_1^2+nx_2^2+...+nx_n^2} why is this greater than ||x_1||


    IF we square both of them, I get nx_1^2+nx_2^2+...+nx_n^2 and  (|x_1|+|x_2|+....|x_n|)^2 , how to show ||x_2|| si bigger?
    Cauchy–Schwarz inequality: 1|x_1| + 1|x_2| + \ldots + 1|x_n| \leqslant \sqrt{1^2+1^2+\ldots+1^2}\sqrt{|x_1|^2 +|x_2|^2 + \ldots + |x_n|^2}.
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