TITLE: Showing that $l_2$ norm is smaller than $l_1$
QUESTION [37 upvotes]: How can I show that $L_2\le L_1$
$||x||_1\ge ||x||_2$ 
and also we have that
$\|x\|_2\leq \sqrt m\|x\|_{\infty}$               
regarding the first part, can I say that: 
$$
\sqrt{\sum\limits_{i=1}^n x^2 } \leq {\sum\limits_{i=1}^n {\sqrt x}^2 }
$$

REPLY [3 votes]: This is a short proof to show that any vector norm is less than or equal to 1-norm.
According to Minkowski Inequality,
$$\|f+g\|_p\le\|f\|_p+\|g\|_p$$
$\|a\|_p=$($\sum_{i=0}^n |a_i|^p)^{1/p}\le (\sum_{i=0}^{n-1} |a_i|^p)^{1/p} +(|a_n|^p)^{1/p}\\\le (\sum_{i=0}^{n-2} |a_i|^p)^{1/p} +(|a_{n-1}|^p)^{1/p} + (|a_n|^p)^{1/p} \\ ...\\...\\ \le(|a_1|^p)^{1/p} +(|a_2|^p)^{1/p} +....+ (|a_n|^p)^{1/p}\\=|a_1| +|a_2| +....+ |a_n| = \|a\|_1 \\ $
$\implies\|a\|_p\le \|a\|_1$ for any $ p\gt1$