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.
Best Answer
If I remember correctly, Cauchy proved the inequality that you don't want proven. The generalisation that you do want proven was proven later by Schwarz. So, I think that the former inequality ought to be called the Cauchy inequality. Oh well.
The proof of the (general) Cauchy-Schwarz inequality essentially comes down to orthogonally decomposing $x$ into a component parallel to $y$ and a component perpendicular to $y$, and using the fact that the perpendicular component has a non-negative square norm. That is, we start from the statement: $$\left\|x - \frac{\langle x, y \rangle}{\langle y, y \rangle} y\right\|^2 \ge 0.$$ The rest is just expanding the inner product. We have, \begin{align*} 0 &\le \left\langle x - \frac{\langle x, y \rangle}{\langle y, y \rangle} y, x - \frac{\langle x, y \rangle}{\langle y, y \rangle} y\right\rangle \\ &= \langle x, x \rangle - \frac{\overline{\langle x, y \rangle}}{\langle y, y \rangle} \langle x, y \rangle - \frac{\langle x, y \rangle}{\langle y, y \rangle} \langle y, x \rangle + \frac{\langle x, y \rangle}{\langle y, y \rangle} \frac{\overline{\langle x, y \rangle}}{\langle y, y \rangle} \langle y, y\rangle \\ &= \langle x, x \rangle - 2\frac{|\langle x, y \rangle|^2}{\langle y, y \rangle} + \frac{|\langle x, y \rangle|^2}{\langle y, y \rangle} \\ &= \langle x, x \rangle - \frac{|\langle x, y \rangle|^2}{\langle y, y \rangle}. \end{align*} Rearranging yields the inequality.