Proof that the Euclidean norm is indeed a norm

Let $x=(x_1,x_2,\dots,x_n)\in\mathbb{R}^n$. $cx=(cx_1,cx_2,\dots,cx_n).$ The definition for the euclidean norm is the following:

$||x||=\sqrt{\sum_{i=1}^{n}x_i^2}$

For 2, $||cx||=\sqrt{\sum_{i=1}^{n}(cx_i)^2}=\sqrt{\sum_{i=1}^{n}(c^2x_i^2)}=\sqrt{c^2\sum_{i=1}^{n}x_i^2}=|c|\sqrt{\sum_{i=1}^{n}x_i^2}=c||x||$.

You can consult, for example, this article: https://bspace.berkeley.edu/access/content/group/2fb5bd3e-8d09-40ee-a371-6cc033d854b9/ho4.pdf

I am grateful for the answer to my question and the comments. However, as my title says, this is a question about Euclidean norm being a norm. So for completeness I will provide the proof for 3 (sub-additivity). My proof goes like this:

$$\| x + y\|^2 = \sum_{i=1}^{n}(x_i+y_i)^2 = \sum_{i=1}^{n}(x_i^2+y_i^2+2x_iy_i) = \sum_{i=1}^{n}x_i^2+ \sum_{i=1}^{n} y_i^2+ 2\sum_{i=1}^{n}x_iy_i $$

Thus:

$$ \| x + y\|^2 = \| x \|^2 + \| y\|^2 + 2 x \cdot y $$

Now lets use Cauchy-Schwarz Inequality (as suggested by jimbo in the comments), we get:

$$ \| x + y\|^2 = \| x \|^2 + \| y\|^2 + 2 x \cdot y \leq \| x \|^2 + \| y\|^2 + 2 \| x \| \| y \| = ( \| x \| + \| y \|)^2$$

which is the same as:

$$\| x + y\|^2 \leq ( \| x \| + \| y \|)^2$$

which implies:

$$ \| x + y\| \leq \| x \| + \| y \|$$

as required.

Thnx everyone! :)