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 ?
Best Answer
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.