Showing that $l_2$ norm is smaller than $l_1$ 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: 
$$
\sqrt{\sum\limits_{i=1}^n x^2 } \leq {\sum\limits_{i=1}^n {\sqrt x}^2 }
$$ 
 A: I assume you are using finite dimensional vector spaces (looks like a familiar question from Golub and Van Loan).
\begin{align}
||x||_2^{2}=\sum_{i=1}^{N}|x_i|^2\leq\left(\sum_{i=1}^{N}|x_i|^2+2\cdot\sum_{i,j,i< 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\cdot\max_{i}(|x_i|^2)=N||x||_{\infty}^{2}
\end{align}
This implies $||x||_2\leq \sqrt{N}||x||_{\infty}$
A: 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$
A: 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.
A: Here is a proof that works for any norm on a finite dimensional vector space, not just $\ell^{N}_{p}$ norms for $p \in [1, \infty)$.
Assume wlog that $\{ \mathbf{e}_i\}_{i=1}^N$ is a basis such that $\| \mathbf{e}_i\| = 1$. (if not you can choose a new basis by normalizing the basis vectors.) In $\mathbb R^N$, you can choose the unit vectors along coordinate axes.
Now,
\begin{align*}
\| x \| &= \left\| \sum_{i=1}^{N} x_i \mathbf{e}_i \right\|   \\
&\le \sum_{i=1}^{N} |x_i| \quad \left( \text{by triangle inequality and } \|\mathbf{e_i}\| = 1 \right) \\
&= \|x\|_1
\end{align*}
A: Hint for Squirtle's generalization: calculate derivative of $f(x)=(a_1^{x}+a_2^x+...+a_n^x)^{1/x}$ and prove $f'(x)<0,\forall x\geq 1$
