Proving that for all $x\in\mathbb{R}^n$, $\|x\|_\infty \le \|x\| \le \sqrt{n}\,\|x\|_\infty$ I believe I proved this incorrectly, can someone please help me with validly proving the problem below? 
$\def\x{{\bf x}}$
Define the $\infty$-norm on $\mathbb{R}^n$ by 
$$\|\x\|_\infty = \|(x_1,x_2,\dots,x_n)\|_\infty = \max_{1\le j\le n} |x_j|.$$
Prove that for all $\x\in\mathbb{R}^n$, 
$$\|\x\|_\infty \le \|\x\| \le \sqrt{n}\,\|\x\|_\infty$$
where $\|\x\|=\|(x_1,x_2,\dots,x_n)\| = \bigg(\sum_{j=1}^n |x_j|^2\bigg)^{1/2}$ is the
usual Euclidean norm (also denoted by $\|\x\|_2$).
Proof: Let $|\x_t| = \max\{|\x_1|, |\x_2|, \dots, |\x_n|\}$ and $||\x||_{\infty} \le (|\x_1|^2 + |\x_2|^2 + \dots + |\x_n|^2)^{\frac{1}{2}}$. Since, $$||\x|| =  (|\x_1|^2 + |\x_2|^2 + \dots + |\x_n|^2)^{\frac{1}{2}},$$ therefore, $$||\x||_{\infty} \le ||\x|| = (|\x_1|^2 + |\x_2|^2 + \dots + |\x_n|^2)^{\frac{1}{2}}$$ $$= |\x_t|(|\frac{\x_1}{\x_t}|^2 + |\frac{\x_2}{\x_t}|^2 + \dots + |\frac{\x_n}{\x_t}|^2)^{\frac{1}{2}} \le n^{\frac{1}{2}}||\x||_{\infty}$$. 
Since $||\x||_{\infty} = |\x_t|$ and $|\frac{\x_i}{\x_t}| < 1$ for all $i = 1(1)n.$ So, the result $||\x||_{\infty} \le ||\x|| \le \sqrt{n}||\x||_{\infty}$ is complete and we are done. 
 A: $\def\x{{\bf x}}$
$\textbf{Solution:}$ We know if $\x = (\x_1, \x_2, \dots, x_n) \in \mathbb{R}^n$, then $||\x||= \bigg(\sum_{j=1}^n |x_j|^2\bigg)^{1/2}$ and $||\x||_{\infty} = \sup\{||\x_j: j =1,2,\dots, n\}$. Now, $|\x_i| \le \sup\{||\x_j: j =1,2,\dots, n\}$ for all $i$ implies $|\x_i|\le ||\x||_\infty$ for all $j = 1,2,\dots, n$. So, $|\x_j|^2 \le (||\x||_{\infty})^2$ and implies $\x_j^2 \le (||\x||_{\infty})^2$ [*].
Now, $$||\x|| = \bigg(\sum_{j=1}^n |\x_j|^2\bigg)^{1/2} \le \bigg(\sum_{j=1}^n (||\x_j||_{\infty})^2\bigg)^{1/2}$$ by [*] which is equal to $$(||\x||_{\infty}^2 + ||\x||_{\infty}^2 +\dots + ||\x||_{\infty}^2)^{1/2}$$ (n-times) $$=(n||\x||_{\infty}^2)^{1/2} = \sqrt{n} ||\x||_{\infty}.$$
So $||\x|| \le \sqrt{n} ||\x||_{\infty}.$ [**]
Now, let $\sup\{|\x_j| : j =1,2,\dots, n\} = |\x_{j_0}|.$ Then, $$|\x_{j_0}| = (|\x_{j_0}|^2)^{1/2} \le (|\x_1|^2 + |\x_2|^2+\dots +|\x_{j_0}|^2 + \dots + |\x_n|^2)^{1/2}$$ $$= \bigg(\sum_{j=1}^n |\x_j|^2\bigg)^{1/2} = ||\x||.$$ 
So, $|\x_{j_0}| \le ||\x||$ and $\sup\{|\x_j| : j =1,2,\dots, n\} \le ||\x||$ implies $||\x||_{\infty} \le ||\x||$ [**]. 
From [*] and [**] we get $||\x||_{\infty} \le ||\x|| \le \sqrt{n}||\x||_{\infty}$. 
