TITLE: Cauchy-Schwarz inequality proof (but not the usual one)
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: Here is the proof (over the field of complex numbers) of the additional claim that you mentioned. Cauchy-Schwarz inequality is also proven.
Problem: Show that for an inner product space $V$ over $\mathbb{C}$
$$|\langle x,y \rangle|=\|x\| \cdot \|y\| \quad \text{if and only if} \quad x=\lambda y, \, \text{where} \, \lambda \in \mathbb{C}.$$
Solution: Notice that substituting $\lambda y$ for $x$ gives us
\begin{equation}
\begin{split}
|\langle x,y \rangle |
 & = |\langle \lambda y, y \rangle |\\
 & = |\lambda \cdot \langle y,y \rangle |\\
 & = | \lambda | \cdot \|y\| \cdot \|y\|\\
 & = \|\lambda y\| \cdot \|y\|\\
 & = \|x\| \cdot \|y\|\\
\end{split}
\end{equation}
Conversely, assume that $|\langle x,y \rangle|=\|x\| \cdot \|y\|$ for some $x,y \in V$. Then notice that by the property of an inner product, $\forall \, \lambda \in \mathbb{C}$
$$\langle x-\lambda y, x-\lambda y \rangle \geq 0$$
Also, notice that the left side can be rewritten as
\begin{equation}
\begin{split}
\langle x-\lambda y, x-\lambda y \rangle
 & = \langle x,x \rangle - \lambda \langle y,x \rangle - \bar{\lambda} \langle x,y \rangle + \lambda \bar{\lambda} \langle y,y \rangle\\
 & = \langle x,x \rangle - \lambda \overline{\langle {x,y} \rangle} - \bar{\lambda} \langle x,y \rangle + \lambda \bar{\lambda} \langle y,y \rangle\\
 & = \langle x,x \rangle - 2\text{Re} \left(\lambda \overline{\langle {x,y} \rangle}\right) +|\lambda|^2 \langle y,y \rangle\\
\end{split}
\end{equation}
We know that $\langle x,y \rangle = re^{i\theta}$, where $r=|\langle x,y \rangle|$ and $\theta$ is some angle. Then $\lambda \overline{\langle {x,y} \rangle}=\lambda r e^{-i\theta}$. Since this holds for all $\lambda$, Let $\lambda=te^{i \theta}$, where $t$ is any real number. Then 
$$\lambda \overline{\langle {x,y} \rangle}=te^{i \theta} re^{-i \theta}=tr \in \mathbb{R}$$
Thus, for this  $\lambda$ (where $t$ can be any real number), the above equation and equality can be combined to give us 
\begin{equation}
\begin{split}
\langle x-\lambda y, x-\lambda y \rangle
 & = \langle x,x \rangle - \lambda \langle y,x \rangle - \bar{\lambda} \langle x,y \rangle + \lambda \bar{\lambda} \langle y,y \rangle\\
 & = \langle x,x \rangle - \lambda \overline{\langle {x,y} \rangle} - \bar{\lambda} \langle x,y \rangle + \lambda \bar{\lambda} \langle y,y \rangle\\
 & = \langle x,x \rangle - 2\text{Re} \left(\lambda \overline{\langle {x,y} \rangle}\right) +|\lambda|^2 \langle y,y \rangle\\
 & = \langle x,x \rangle -2tr +t^2 \langle y,y \rangle\\
 & \geq 0
\end{split}
\end{equation}
Since this expression is positive for all $t$ and since the above inequality is a quadratic function of $t$, we know that its discriminant has to be smaller or equal $0$. Thus,
$$4r^2-4 \langle x,x \rangle \cdot \langle y,y \rangle \leq 0$$
Plugging back in $|\langle x,y \rangle|$ for $r$ gives us
$$|\langle x,y \rangle|^2\leq \langle x,x \rangle \cdot \langle y,y \rangle$$
which is equivalent to 
$$|\langle x,y \rangle| \leq \|x\| \cdot \|y\|$$
which is known as the Cauchy-Schwarz inequality.
However, we know that $|\langle x,y \rangle|=\|x\| \cdot \|y\|$. Thus, it has to be that 
$$4r^2-4 \langle x,x \rangle \cdot \langle y,y \rangle = 0$$
But discriminant being $0$ implies that there exists $t$ such that 
$$\langle x,x \rangle -2tr +|\lambda|^2 \langle y,y \rangle=0$$
Choose that $t$. Then for $\lambda=te^{i \theta}$, where $t$ is the specific value that we just chose, we get
\begin{equation}
\begin{split}
\langle x-\lambda y, x-\lambda y \rangle
 & = \langle x,x \rangle - \lambda \langle y,x \rangle - \bar{\lambda} \langle x,y \rangle + \lambda \bar{\lambda} \langle y,y \rangle\\
 & = \langle x,x \rangle - \lambda \overline{\langle {x,y} \rangle} - \bar{\lambda} \langle x,y \rangle + \lambda \bar{\lambda} \langle y,y \rangle\\
 & = \langle x,x \rangle - 2\text{Re} \left(\lambda \overline{\langle {x,y} \rangle}\right) +|\lambda|^2 \langle y,y \rangle\\
 & = \langle x,x \rangle -2tr +t^2 \langle y,y \rangle\\
 & = 0
\end{split}
\end{equation}
But $\langle x-\lambda y, x-\lambda y \rangle=0$ implies that $x-\lambda y=0$, which in turns gives us the desired result, namely $x=\lambda y$.