Consider the following variance:
\[ \operatorname{Var}\left[ \frac{X}{\sigma_X} ± \frac{Y}{\sigma_Y} \right] \]
Using this property:
\begin{align} \operatorname{Var}\left[ \frac{X}{\sigma_X} ± \frac{Y}{\sigma_Y} \right] &= \frac{1}{\sigma_X^2} \operatorname{Var}[X] + \frac{1}{\sigma_Y^2}\operatorname{Var}[Y] ± \frac{2}{\sigma_X^2 \sigma_Y^2}\operatorname{cov}(X,Y) \\ &= \frac{1}{\sigma_X^2} \sigma_X^2 + \frac{1}{\sigma_Y^2} \sigma_Y^2 ± \frac{2}{\sigma_X^2 \sigma_Y^2}\operatorname{cov}(X,Y) \\ &= 2 ± \frac{2}{\sigma_X^2 \sigma_Y^2}\operatorname{cov}(X,Y) \end{align}
Since variance is always \(\ge 0\):
\begin{gather} \operatorname{Var}\left[ \frac{X}{\sigma_X} ± \frac{Y}{\sigma_Y} \right] \ge 0 \\ 0 \le 2 ± \frac{2}{\sigma_X^2 \sigma_Y^2}\operatorname{cov}(X,Y) \end{gather}
Dividing both sides by 2:
\begin{gather} 0 \le 1 ± \frac{\operatorname{cov}(X,Y)}{\sigma_X^2 \sigma_Y^2} \\ -1 \le ± \frac{\operatorname{cov}(X,Y)}{\sigma_X^2 \sigma_Y^2} \end{gather}
If \(\frac{\operatorname{cov}(X,Y)}{\sigma_X^2 \sigma_Y^2}\) is positive:
\[ -1 \le \frac{\operatorname{cov}(X,Y)}{\sigma_X^2 \sigma_Y^2} \]
If \(\frac{\operatorname{cov}(X,Y)}{\sigma_X^2 \sigma_Y^2}\) is negative:
\begin{gather} -1 \le -\frac{\operatorname{cov}(X,Y)}{\sigma_X^2 \sigma_Y^2} \\ 1 \ge \frac{\operatorname{cov}(X,Y)}{\sigma_X^2 \sigma_Y^2} \end{gather}
This means:
\[ -1 \le \frac{\operatorname{cov}(X,Y)}{\sigma_X^2 \sigma_Y^2} \le 1 \]