Prove $\|x\|\le\sum_{i=1}^n|x_i|\le\sqrt{n}\|x\|$ for all $x\in\mathbb{R}^n$. Prove $\|x\|\le\sum_{i=1}^n|x_i|\le\sqrt{n}\|x\|$ for all $x\in\mathbb{R}^n$.
I'm stumped, can't find any intuition here or where to begin. Any suggestions?
 A: One direction is really easy: if $x=(x_1,\dots,x_n)$, you have:
$$\sum_{i=1}^nx_i^2\leq\left(\sum_{i=1}^n|x_i|\right)^2$$
because the right-hand-side will have the sum of squares plus other non-negative terms. Taking the square root gives the left-hand-side.
As for the right-hand-side, namely $\sum_{i=1}^n|x_i|\leq\sqrt{n}||x||$,
You may assume $x_i\neq 0$ for all $i$, otherwise just delete the zeros from both sides. Choose $\epsilon_i=+1$ if $x_i>0$ and $\epsilon_i=-1$ if $x_i<0$. Then $\sum_{i=1}^n|x_i|=\sum_{i=1}^n\epsilon_ix_i$. This expression is simply the inner product of the vectors $(x_1,\dots,x_n)$ and $(\epsilon_1,\dots,\epsilon_n)$. The Cauchy-Schwartz-Bunyakovski inequality states that for any two vectors $x,y\in \mathbb{R}^n$, the absolute value of their inner product is at most the product of their norms:
$$|\langle x,y\rangle|\leq ||x||||y||$$
applying this to our situation, and noting that the (Euclidean) norm of any vector whose components are $\pm 1$ is $\sqrt{n}$, we obtain the r.h.s inequality.
A: Here's the intuition: 


*

*$||x||=r$ is the sphere of radius $r$ centered on the origin in $\mathbb{R}^n$.

*$\sum_{i=1}^n|x_i| = r$ is the "diamond" centered on the origin with vertices at the unit points on the axes. In 2-D this diamond is a square, in 3-D it is an octahedron, and I don't know the higher dimensional names.


So, this inequality is all about how to inscribe diamonds inside spheres and how to circumscribe diamonds outside of spheres.
