Proof of $\|x\|_1\le \sqrt{n}\cdot\|x\|_2$

Question

How can I prove that $\|x\|_1\le \sqrt{n}\cdot\|x\|_2$?

What I tried: $\|x\|_1=\sum |x_i|=\sum\sqrt{|x_i|^2}$

$x=(x_1,x_2,...,x_n)$

Answer 1

Hint Use the Cauchy–Schwarz inequality

Answer 2

Just to add a proof here as suggested by N.S: pick $y = (y_i)$ with $|y_i| = 1$ and $y_i \overline{x_i}= |x_i|$.

From Cauchy Schwarz (which holds for Hermitian dot product)

$$(\sum\limits_{i=1}^n |x_i|^2) = |x^H y| \leq \langle x,x \rangle \cdot \langle y,y \rangle = n \sum\limits_{i=1}^n |x_i|^2$$

From which follows $||x ||_1 \leq \sqrt{n}||x ||_2 $