Proof of Cauchy Schwarz Inequality Let $V$ be a vector space over either $\mathbb{R}$ or $\mathbb{C}$
Let $\langle.,.\rangle$ be an inner product on $V$.
Prove the Cauchy Schwarz Inequality
$$\forall x,y\in V, |\langle x,y\rangle|^2 \leq |\langle x,x\rangle||\langle y,y\rangle|$$
all the proofs I can come up with right now involves some sort of norm on $V$.
I don't think I can just assume that $V$ has a norm.
So are there ways to prove it using nothing but the definition of inner product?
 A: Let $\;c\in\Bbb R\;$ , so
$$0\le\langle x+cy,\,x+cy\rangle=\langle x,x\rangle+2c\,\text{Re}\,\langle x,y\rangle+c^2\langle y,y\rangle\le\langle x,x\rangle+2c\,|\langle x,y\rangle|+c^2\langle y,y\rangle$$
The above is a non-negative real quadratic in the real variable $\;c\;$ , so its discriminant must be non-positive:
$$\Delta=4|\langle x,y\rangle|^2-4\langle x,x\rangle\langle y,y\rangle\le0\implies|\langle x,y\rangle|^2\le\langle x,x\rangle\langle y,y\rangle$$
A: Let $\alpha=\frac{\langle x,y\rangle}{|\langle x,y\rangle|}$, then
$$
\begin{align}
0
&\le\left\langle\frac{x}{\|x\|}-\alpha\frac{y}{\|y\|},\frac{x}{\|x\|}-\alpha\frac{y}{\|y\|}\right\rangle\\[9pt]
&=\frac{\langle x,x\rangle}{\|x\|^2}+\frac{\langle y,y\rangle}{\|y\|^2}-\frac{2\mathrm{Re}\left(\overline\alpha\langle x,y\rangle\right)}{\|x\|\,\|y\|}\\
&=2-2\mathrm{Re}\left(\frac{\overline{\langle x,y\rangle}}{|\langle x,y\rangle|}\frac{\langle x,y\rangle}{\|x\|\,\|y\|}\right)\\
&=2-2\frac{|\langle x,y\rangle|}{\|x\|\,\|y\|}\\
\end{align}
$$
Therefore,
$$
|\langle x,y\rangle|\le\|x\|\,\|y\|
$$
where $\|x\|^2=\langle x,x\rangle$.
A: Our papers [Novi Sad J. Math. 47(2017), 177--188] and 
[Rostock. Math. Colloq. 71\,(2016), 28--40 offer some 
different proofs. 
