We know that the ratio of the explained variation to the total variation is called the coefficient of determination which is the square of the Correlation Coefficient Range lies between $-1$ and $+1$. This ratio (coefficient of determination) is non-negative, therefore denoted by $r^2$, thus
\begin{align*}
r^2&=\frac{\text{Explained Variation}}{\text{Total Variation}}\\
&=\frac{\sum (\hat{Y}-\overline{Y})^2}{\sum (Y-\overline{Y})^2}
\end{align*}
It can be seen that if the total variation is all explained, the ratio $r^2$ (Coefficient of Determination) is one and if the total variation is all unexplained then the explained variation and the ratio $r^2$ are zero.
The square root of the coefficient of determination is called the correlation coefficient, given by
\begin{align*}
r&=\sqrt{ \frac{\text{Explained Variation}}{\text{Total Variation}} }\\
&=\pm \sqrt{\frac{\sum (\hat{Y}-\overline{Y})^2}{\sum (Y-\overline{Y})^2}}
\end{align*}
and
\[\sum (\hat{Y}-\overline{Y})^2=\sum(Y-\overline{Y})^2-\sum (Y-\hat{Y})^2\]
Therefore
\begin{align*}
r&=\sqrt{ \frac{\sum(Y-\overline{Y})^2-\sum (Y-\hat{Y})^2} {\sum(Y-\overline{Y})^2} }\\
&=\sqrt{1-\frac{\sum (Y-\hat{Y})^2}{\sum(Y-\overline{Y})^2}}\\
&=\sqrt{1-\frac{\text{Unexplained Variation}}{\text{Total Variation}}}=\sqrt{1-\frac{S_{y.x}^2}{s_y^2}}
\end{align*}
where $s_{y.x}^2=\frac{1}{n} \sum (Y-\hat{Y})^2$ and $s_y^2=\frac{1}{n} \sum (Y-\overline{Y})^2$
\begin{align*}
\Rightarrow r^2&=1-\frac{s_{y.x}^2}{s_y^2}\\
\Rightarrow s_{y.x}^2&=s_y^2(1-r^2)
\end{align*}
Since variances are non-negative
\[\frac{s_{y.x}^2}{s_y^2}=1-r^2 \geq 0\]
Solving for inequality we have
\begin{align*}
1-r^2 & \geq 0\\
\Rightarrow r^2 \leq 1\, \text{or}\, |r| &\leq 1\\
\Rightarrow & -1 \leq r\leq 1
\end{align*}
Therefore, the Correlation Coefficient Range lies between $-1$ and $+1$ inclusive.
Alternative Proof: Correlation Coefficient Range
Since $\rho(X,Y)=\rho(X^*,Y^*)$ where $X^*=\frac{X-\mu_X}{\sigma_X}$ and $Y^*=\frac{Y-Y^*}{\sigma_Y}$
and as covariance is bi-linear and $X^*, Y^*$ have zero mean and variance 1, therefore
\begin{align*}
\rho(X^*,Y^*)&=Cov(X^*,Y^*)=Cov\{\frac{X-\mu_X}{\sigma_X},\frac{Y-\mu_Y}{\sigma_Y}\}\\
&=\frac{Cov(X-\mu_X,Y-\mu_Y)}{\sigma_X\sigma_Y}\\
&=\frac{Cov(X,Y)}{\sigma_X \sigma_Y}=\rho(X,Y)
\end{align*}
We also know that the variance of any random variable is $\ge 0$, it could be zero i.e. $(Var(X)=0)$ if and only if $X$ is a constant (almost surely), therefore
\[V(X^* \pm Y^*)=V(X^*)+V(Y^*)\pm2Cov(X^*,Y^*)\]
As $Var(X^*)=1$ and $Var(Y^*)=1$, the above equation would be negative if $Cov(X^*,Y^*)$ is either greater than 1 or less than -1. Hence \[1\geq \rho(X,Y)=\rho(X^*,Y^*)\geq -1\].
If $\rho(X,Y )=Cov(X^*,Y^*)=1$ then $Var(X^*- Y ^*)=0$ making $X^* = Y^*$ almost surely. Similarly, if $\rho(X,Y )=Cov(X^*,Y^*)=-1$ then $X^* = – Y^*$ almost surely. In either case, $Y$ would be a linear function of $X$ almost surely.
For proof of Cauchy-Schwarz Inequality please follow the link
We can see that the Correlation Coefficient range lies between $-1$ and $+1$.
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