# show that the inequality is true

• Oct 17th 2011, 04:05 PM
wopashui
show that the inequality is true
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?
• Oct 18th 2011, 12:04 AM
Opalg
Re: show that the inequality is true
Quote:

Originally Posted by 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?

Cauchy–Schwarz inequality: $\displaystyle 1|x_1| + 1|x_2| + \ldots + 1|x_n| \leqslant \sqrt{1^2+1^2+\ldots+1^2}\sqrt{|x_1|^2 +|x_2|^2 + \ldots + |x_n|^2}.$