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. 

Since $\langle tx+y,tx+y\rangle\ge0$ for all $t$, $\;\;\langle x,x\rangle t^2+2\langle x,y\rangle t+\langle y, y\rangle\ge0$ for all $t$, so
$(2\langle x,y\rangle)^2-4\langle x,x\rangle\langle y,y\rangle\le0\implies \langle x,y\rangle^2\le\langle x,x\rangle\langle y,y\rangle=\lvert\lvert x\rvert\rvert^2 \lvert\lvert y\rvert\rvert^2$.
