# 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 ?

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).
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*}