show that the inequality is true

**wopashui** · October 17th 2011, 04:05 PM

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?

**Opalg** · October 18th 2011, 12:04 AM

Originally Posted by wopashui

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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