Confusion - Correlation between -1 and 1 So in the text book, it has the following proof of correlation being between $-1$ and $1$:

This proof seems solid, but the only thing I am not clear about is: 
Why is the variance of the ratio $X$ and $\sigma$, i.e., $Var(X/\sigma$) equal to $Var(X)/\sigma^2_x$?
 A: The variance is the square of the deviation. So $\text{Var } x=\sigma_x^2$ and then the numerator and denominator are equal.
A: Definition:
$$\operatorname{Var}(X) = \frac 1 n \sigma_X^2= \sum (X_i - \bar X)^2$$
$$\operatorname{Var} \left(\frac {X}{\sigma_X}\right) = $$
$$\frac 1 n \sum \left(\frac {X_i}{\sigma_X} - \frac {\bar X}{\sigma_X}\right)^2\\
\frac 1 n \sum \left(\frac {X_i-\bar X}{\sigma_X}\right)^2\\
\frac 1{\sigma_X^2} \cdot \frac 1 n \sum (X_i-\bar X)^2\\
\frac 1{\sigma_X^2} \operatorname{Var}(X)=\frac 1{\sigma_X^2} \sigma_X^2 = 1$$
Definition:
$$\operatorname{Cov}(X,Y) = \frac 1 n \sum (X_i - \bar X)(Y_i - \bar Y)$$
The Cauchy-Schwartz inequality:
$$\left(\sum uv\right)^2 \le \sum u^2\sum v^2$$
Substitute $u = (X_i - \bar X)$ and $v = (Y_i - \bar Y)$
$$\left(\sum (X_i - \bar X)(Y_i - \bar Y)\right)^2 \le \sum (X_i - \bar X)^2\sum (Y_i - \bar Y)^2\\
|\operatorname{Cov}(X,Y)| \le \sigma_X\sigma_Y$$
Definition:
$$\rho_{X,Y} = \frac {\operatorname{Cov}(X,Y)}{\sigma_X\sigma_Y}$$
$$|\operatorname{Cov}(X,Y)|\le\sigma_X\sigma_Y \\
|\rho_{X,Y}| \le 1$$
And it might be worth noting that $\sigma_X, \sigma_Y$ are equivalent to distance measures, and $\rho_{X,Y}$ is equivalent to $\cos \theta$
Law of cosines:
$$c^2 = a^2 + b^2 + 2ab\cos \theta$$
Has the parallel:
$$\sigma_{X+Y}^2 = \sigma_X^2 + \sigma_Y^2 + 2\sigma_X\sigma_Y \rho_{X,Y}$$
