What is the relationship between the Cauchy–Schwarz Inequality and angles in $R^n$?

Question

What is the relationship between the Cauchy–Schwarz Inequality and angles in $R^n$?

Answer 1

The angle $\theta$ between two nonzero vectors $u$ and $v$ is given by: $$\theta=\cos ^{-1}\left(\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}\right)$$ SInce the value of $\theta$ depends on the cosine function function then it implies that: $$-1 \leq \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|} \leq 1$$ The Cauchy-Schwarz Inequality states that: If $\mathbf{u}=\left(u_{1}, u_{2}, \ldots, u_{n}\right)$ and $\mathbf{v}=\left(v_{1}, v_{2}, \ldots, v_{n}\right)$ are vectors in $R^{n}$, then $$ |\mathbf{u} \cdot \mathbf{v}| \leq\|\mathbf{u}\|\|\mathbf{v}\| $$ or in terms of components $$ \left|u_{1} v_{1}+u_{2} v_{2}+\cdots+u_{n} v_{n}\right| \leq\left(u_{1}^{2}+u_{2}^{2}+\cdots+u_{n}^{2}\right)^{1 / 2}\left(v_{1}^{2}+v_{2}^{2}+\cdots+v_{n}^{2}\right)^{1 / 2} $$