Absolute value of vector components When is it possible to assume that for any vector $\vec{x}=(x_1,\dots, x_n)\in\mathbb{R}^n$;
$$|x_{i}| \leq \|\vec{x}\|$$
For the case in  $\mathbb{R}^2$:
$$|x_2| \leq \|\vec{x}\|$$
$$\left|x_2\right| \leq x_2 \cdot \sqrt{\left(\frac{x_1}{x_2}\right)^2+1}$$
Define $\alpha = \sqrt{\left(\frac{x_{1}}{x_2}\right)^2+1}$
Then $|x_i| \leq \|\vec{x}\| \iff \alpha >1$.
Is there a more general result ? Does this reasoning hold ?
 A: Let $X=\mathbb{R}^{N}$. Then, for $x\in X$, the (square of the) $2$-norm is defined as:
\begin{equation*} \|x\|_{2}^{2}=\sum_{i=1}^{N}|x_{i}|^{2}\geq |x_{i}|^{2} \text{ for a given index } i \qquad x=(x_{1},\ldots,x_{N}) \end{equation*}
so taking the square root gives $\|x\|_{2}\geq |x_{i}|$ for all $i\in\{1,\ldots,N\}$. Next, it is a standard result that every norm on a finite-dimensional space (like $X$ in this case) is equivalent, so given another norm, say $\|\cdot\|'$, there are constants $c_{1},c_{2}>0$ such that
\begin{equation*} c_{1}\|x\|'\leq\|x\|_{2}\leq c_{2}\|x\|' \end{equation*}
for all $x\in X$. In particular, it would follow that
\begin{equation*} |x_{i}|\leq\|x\|_{2}\leq c_{2}\|x\|' \end{equation*}
so you can bound individual components by the overall norm for any norm on $X$, up to a constant dependent on the norm (with this constant being $1$ for the $2$-norm).
A: Assuming the notation $ \Vert \vec x \Vert $ denotes the standard norm in $ \mathbb R^n $, the answer to your first question is always. The argument is as follows. Let $ x $ be in $ \mathbb R^n $ and $ x_i $ be its $i^{th} $ component. Then,
\begin{align*}
     | x_i |
          & = \sqrt{ |x_i|^2 } 
          \\
          & \le  \sqrt{ |x_i|^2 + \sum_{j \neq i} |x_j|^2} 
          \\
          & = \Vert \vec x \Vert
          .
\end{align*}
The key step here is the inequality above, which relies on the fact that the function $ f(t) = \sqrt{t} $ is increasing.
The argument you provide is a good place to start for the $ \mathbb R^2 $ case, but it requires some fixes: (1) exclude the $ x_2 = 0 $ case in order to divide by $ x_2 $, (2) replace the factor $ x_2 $ with $|x_2|$ on the right (remember: $ \sqrt{z^2} = |z| $ for real $ z $), (3) replace $ \alpha > 1 $ with $ \alpha \ge 1 $, and (4) state the connection between the statements you provide. After implementing these fixes, you can then establish that the inequality is always true in $ \mathbb R^2 $ by observing that $ \alpha \ge 1 $ always holds. This final bit follows also from the fact that $ f(t) = \sqrt{t} $ is increasing:
\begin{align*}
     \alpha
          & = f\left( 1 + \left(\frac{x_1}{x_2}\right)^2 \right)
          \\
          & \ge  f(1)
          \\
          & = 1
          .
\end{align*}
