TITLE: 1 and 2 norm inequality
QUESTION [17 upvotes]: While looking over my notes, my lecturer stated the following inequality; $$\|x\|_2 \leq \|x\|_1 \leq \sqrt{n}\|x\|_2$$ where $x \in \mathbb{R^n}.$ There was no proof given, and I've been trying to prove it for a while now. I know the definitions of the $1$ and $2$ norm, and, numerically the inequality seems obvious, although I don't know where to start rigorously. 
Thank you.

REPLY [2 votes]: Another way to show that $||x||_1 \leq \sqrt{n}||x||_2$ is as follows:
\begin{align*}
\begin{array}{c}
\displaystyle 0\leq \sum_{i\neq j}\big(|x_i|-|x_j|\big)^2 = (n-1)\sum_{i=1}^{n} |x_i|^2 - 2\sum_{i\neq j}|x_i||x_j|  \\
\displaystyle\Rightarrow\quad  2\sum_{i\neq j}|x_i||x_j|\leq (n-1)\sum_{i=1}^{n} |x_i|^2 \\
\displaystyle\Rightarrow\quad \sum_{i=1}^{n} |x_i|^2 + 2\sum_{i\neq j}|x_i||x_j|\leq n\sum_{i=1}^{n} |x_i|^2  \\
\displaystyle\Rightarrow\quad  \left( \sum_{i=1}^n |x_i| \right)^2 \leq n\sum_{i=1}^{n} |x_i|^2 \\
\displaystyle\Rightarrow \quad ||x||_1\leq \sqrt{n}||x||_2
\end{array}
\end{align*}