TITLE: Why is this true: $\|x\|_1 \le \sqrt n \cdot \|x\|_2$?
QUESTION [19 upvotes]: Stuck, help please, I am really new to this.
I opened the 2-norm and multiplied by $n$, then I am thinking to square both sides.
The problem is that I do not know how to open $(x_1 + x_2 + ... + x_n)^2$
$\|x\|_1 \leqslant \sqrt n\cdot \|x\|_2$

REPLY [23 votes]: Using Cauchy-Schwarz, we get:
$$
\Vert x\Vert_1= 
\sum\limits_{i=1}^n|x_i|=
\sum\limits_{i=1}^n|x_i|\cdot 1\leq
\left(\sum\limits_{i=1}^n|x_i|^2\right)^{1/2}\left(\sum\limits_{i=1}^n 1^2\right)^{1/2}=
\sqrt{n}\Vert x\Vert_2
$$