prove that coefficient of correlation always lies between -1 to +1

coefficient of correlation = r = $\frac{cov(x,y)}{{\sigma }_x{\sigma }_y}$
${{\sigma }_x}^2= \frac{\sum (x-\overline{x}{)}^2}{n}$

${{\sigma }_y}^2= \frac{\sum (y-\overline{y}{)}^2}{n}$

cov(x,y) = $\frac{\sum (x-\overline{x}\left)\right(y-\overline{y})}{n}$

we have to prove $-1\le r\le 1$
let us consider two terms $\left[\frac{x-\overline{x}}{{\sigma }_x}\right]$ and $\left[\frac{y-\overline{y}}{{\sigma }_y}\right]$

take sum of the square of these two variables
$\sum {\left[\left[\frac{x-\overline{x}}{{\sigma }_x}\right]\pm \left[\frac{y-\overline{y}}{{\sigma }_y}\right]\right]}^2$
since it is a square term then it is surely a positive number
that is $\sum {\left[\left[\frac{x-\overline{x}}{{\sigma }_x}\right]\pm \left[\frac{y-\overline{y}}{{\sigma }_y}\right]\right]}^2\ge 0$
opening its square we get
$\sum \left[{\left[\frac{x-\overline{x}}{{\sigma }_x}\right]}^2+{\left[\frac{y-\overline{y}}{{\sigma }_y}\right]}^2+ 2\left[\frac{x-\overline{x}}{{\sigma }_x}\right]\left[\frac{y-\overline{y}}{{\sigma }_y}\right]\right]\ge 0$
on further simplifying it, we get
​$\frac{\sum (x-\overline{x}{)}^2}{{{\sigma }^2}_x}+\frac{\sum (y-\overline{y}{)}^2}{{{\sigma }^2}_y}\pm 2\frac{\sum (x-\overline{x}\left)\right(y-\overline{y)}}{{\sigma }_x{\sigma }_y}\ge 0$
divide the entire expression by n
$\frac{\sum (x-\overline{x}{)}^2}{n{{\sigma }^2}_x}+\frac{\sum (y-\overline{y}{)}^2}{n{{\sigma }^2}_y}\pm 2\frac{\sum (x-\overline{x}\left)\right(y-\overline{y)}}{n{\sigma }_x{\sigma }_y}\ge 0$
using the formula of variance of x, variance of y and covariance of x and y, we get
$\frac{{{\sigma }^2}_x}{{{\sigma }^2}_x}+\frac{{{\sigma }^2}_y}{{{\sigma }^2}_y}\pm 2\frac{cov(x,y)}{{\sigma }_x{\sigma }_y}\ge 0$

1+1$\pm$2r$\ge 0$

$$\begin{gathered} 2+2r\ge 0 \\ 2\ge -2r \\ 1\ge -r \\ r\ge -1 \\ or \\ 2-2r\ge 0 \\ 2\ge 2r \\ 1\ge r \\ therefore we conclude that \\ -1\le r\le 1 \\ always. \end{gathered}$$