Proof $\Vert x\Vert_\infty \leq \Vert x \Vert \leq \sqrt{n}\Vert x\Vert_\infty$

Question

For $x \in \mathbb{R}^n$ we define $\Vert x \Vert _\infty := \sup_{k = 1,..,n} |x_k|$ (meaning that $\Vert x \Vert _\infty $ is the biggest component of $x$ according to amount)

How can one prove that

$$\Vert x\Vert_\infty \leq \Vert x \Vert \leq \sqrt{n}\Vert x\Vert_\infty$$

I have seen it on Wikipedia, but there's no proof to it.

I know that using Cauchy–Schwarz inequality we get for all $x\in\mathbb{R}^n$ $$ \Vert x\Vert_1= \sum\limits_{i=1}^n|x_i|= \sum\limits_{i=1}^n|x_i|\cdot 1\leq \left(\sum\limits_{i=1}^n|x_i|^2\right)^{1/2}\left(\sum\limits_{i=1}^n 1^2\right)^{1/2}= \sqrt{n}\Vert x\Vert_2 $$

but that doesn't help me.

Answer 1

$$\| x \|_\infty = \sup_{k = 1,..,n} |x_k| =\sqrt{ \sup_{k = 1,..,n} |x_k|^2 } \leq \sqrt{ \sum_{k = 1}^n |x_k|^2 }$$

For the second $$\sqrt{ \sum_{j = 1}^n |x_j|^2 } \leq \sqrt{ \sum_{j = 1}^n \sup_{k = 1,..,n} |x_k|^2 }$$

Answer 2

Let $\Vert x \Vert_\infty=l=\sup_\limits{i\in{1,2,...n}}\vert x_i \vert$

Note that $0 \leq l, \Vert x \Vert _2$.

Then $l^2 \leq \sum \limits_{i=1}^n \vert x_i \vert^2 \implies \Vert x \Vert_\infty=l\leq \sqrt{\sum \limits_{i=1}^n \vert x_i \vert^2} =\Vert x\Vert _2$

Also $\sum \limits_{i=1}^n \vert x_i \vert^2 \leq \sum \limits_{i=1}^nl^2 \implies \sqrt{\sum \limits_{i=1}^n \vert x_i \vert^2} \leq l\sqrt n \implies \Vert x \Vert _2 \leq \sqrt n\Vert x \Vert_\infty $