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**wopashui** show that $\displaystyle ||x||_1 <= \sqrt{n}||x||_2$

I have proved that $\displaystyle ||x_2||<=||x_1||$

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

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

IF we square both of them, I get $\displaystyle nx_1^2+nx_2^2+...+nx_n^2$ and $\displaystyle (|x_1|+|x_2|+....|x_n|)^2$ , how to show $\displaystyle ||x_2||$ si bigger?