Cauchy-Schwarz inequality proof (but not the usual one) Before you downvote/vote-to-close, I am not asking for a proof of:
$$\sum^n_{i=1}a_ib_i\le\sqrt{\sum_{i=1}^na_i^2}\sqrt{\sum^n_{i=1}b_i^2 } $$
Which is what EVERY link I've found assumes is the inequality (for a proof of that see: http://www.maths.kisogo.com/index.php?title=Cauchy-Schwarz_inequality&oldid=692 )

I am reading a book that claims the Cauchy-Schwarz inequality is actually:
$$\vert\langle x,y\rangle\vert\le\Vert x\Vert\Vert y\Vert$$ where $\Vert x\Vert :=\sqrt{\langle x,x\rangle}$
with the additional claim: equality holds $\iff\ x,y$ are linearly dependent
I cannot find a proof of this claim (only proofs for the dot product inner product). My question is: what is the simplest way to prove this (I define simplest to mean "using as few definitions outside of the inner product itself as possible" - so that's not including Gramm-Schmitt orthonormalisation, for example.)
I've just found some possible answers in the links on the right (annoyingly) but fortunately I have a second, albeit softer, question:
What inequalities that are common for say $\mathbb{R}^n$ are actually based on the inner product? 
(In future I will check this site first, I was targeting lecture notes and such in my search)
Addendum:
I want to use this on not-finite dimensional vector spaces. I've found one proof that relies on finite-ness. 

It think the usual formulations and proofs of the Cauchy-Schwarz inequality do include statement and proof of the additional claim that equality holds if and only if the vectors are linearly dependent. For the most obvious online example consider the proof from Wikipedia: the before-last sentence says "Moreover, if the relation '$\geq$'... is actually an equality, then... establishes a relation of linear dependence" (this assumes without mentioning the easy converse that in case of linear dependence, the inequality will be an equality; this is just a direct application of basic properties of the scalar product). Note that the mentioned proof does not assume finite dimensionality.
