Confused about Euclidean Norm I am trying to understand that the Euclidean norm $\|x\|_2 = \left(\sum|x_i|^2\right)^{1/2}$ is in fact a norm and having trouble with the triangle inequality.
All the proofs I have referred to involve the Cauchy-Schwarz inequality. But it seems that this inequality is proved in an inner product space, which has additional properties to a normed space.
So, my question is whether starting with any (possibly infinite dimensional) vector space over $\mathbb{C}$ and taking any algebraic basis for it, can the triangle inequality be proved for the Euclidean norm without making assumptions about an inner product or an orthonormal basis ?
(I don't think that infinite dimensionality should be a problem as any two vectors have finite representations in an algebraic basis).

Addendum after 2 answers and comments.
Can one take the Cauchy-Schwarz inequality "out of context" as an algebraic statement about two finite lists $(x_i) $ and $(y_i)$ and then apply it to the complex coefficients of any algebraic basis to say that $\sum \left|x_iy_i^*\right|\leq (\sum|x_i|^2)^{1/2} (\sum|y_i|^2)^{1/2}$ and then complete the proof of the triangle inequality ?
 A: Hint:
Note that the Euclidean norm is a particular case of a $p$-norm and for these norms the triangle inequality can be proved using the Minkowky inequality.
Anyway, the Euclidean norm is the only $p$-norm that satisfies the parallelogram identity ( see: Determining origin of norm), so it is coming  from an inner product.
About the addendum.
In an $n$ dimensional real space we can prove the C-S inequality with simply algebraic methods (see here). So, yes, in this case we can proof the triangle inequality without explicitly using an inner product space.
A: Community wiki: Just to show the C-S proof. We have
$$||x+y||_2^2=\sum|x_i+y_i|^2 = \sum \left|x_i^2+2x_iy_i+y_i^2\right|.$$
Then by the triangle inequality (over $|\cdot|$)
$$\sum \left|x_i^2+2x_iy_i+y_i^2\right|\leq \sum|x_i|^2+2\sum|x_i||y_i|+\sum|y_i|^2.$$
But, by the Cauchy-Schwarz inequality,
$$\sum|x_i|^2+2\sum|x_i||y_i|+\sum|y_i|^2\leq \sum|x_i|^2+2\left(\sum |x_i|^2\right)^{1/2}\left(\sum|y_i|^2\right)^{1/2}+\sum|y_i|^2.$$
But this is just equal to
$$\left(\left(\sum |x_i|^2\right)^{1/2}+\left(\sum |y_i|^2\right)^{1/2}\right)^2=\left(||x||_2+||y||_2\right)^2.$$
Hence,
$$||x+y||_2\leq ||x||_2+||y||_2.$$
Note
Concerning the comments on the triangle inequality using $|\cdot |$. There is a trivial proof: $$|a+b|^2=(a+b)^2=a^2+b^2+2ab=|a|^2+|b|^2+2ab\leq(|a|+|b|)^2\implies|a+b|\leq |a|+|b|.$$ Hence, $$|x^2+2xy+y^2|\leq|x|^2+|2xy+y^2|\leq|x|^2+|2xy|+|y|^2=|x|^2+2|x||y|+|y|^2.$$
