Showing that $l_2$ norm is smaller than $l_1$

I assume you are using finite dimensional vector spaces (looks like a familiar question from golub and loan).

\begin{align} ||x||_2^{2}=\sum_{i=1}^{N}|x_i|^2\leq\left(\sum_{i=1}^{N}|x_i|^2+2*\sum_{i,j,i\neq j}|x_i||x_j|\right)=||x||_1^2 \end{align}

This implies $||x||_2\leq ||x||_1$. Now \begin{align} ||x||_2^{2}=\sum_{i=1}^{N}|x_i|^2\leq N*\max_{i}(|x_i|^2)=N||x||_{\infty}^{2} \end{align} This implies $||x||_2\leq \sqrt{N}||x||_{\infty}$

In fact, we can do something stronger than this

$${\Vert a \Vert}_p = (\sum_{i=0}^n |a_i|^p)^{1/p} \le (\sum_{i=0}^{n-1} |a_i|^p)^{1/p} + |a_n^p|^{1/p} \le \cdots \le \sum_{i=0}^n |a_i| = {\Vert a \Vert}_1$$

Where each inequality is using the Minkowski inequality. Moreover, we can generalize this idea further to show

$${\Vert * \Vert}_q \le {\Vert * \Vert}_p \text{ whenever } p\le q$$

It is a good exercise.