Prove the Cauchy-Schwarz inequality,

$$|\mathbf{x} \cdot \mathbf{y}| \leqslant|\mathbf{x}||\mathbf{y}|$$

for two vectors $\mathbf{x}, \mathbf{y} \in \mathbb{R}^{n}$. Under what condition does equality hold?

Consider a pyramid in $\mathbb{R}^{n}$ with vertices at the origin $O$ and at $\mathbf{e}_{1}, \mathbf{e}_{2}, \ldots, \mathbf{e}_{n}$, where $\mathbf{e}_{1}=(1,0,0, \ldots), \mathbf{e}_{2}=(0,1,0, \ldots)$, and so on. The "base" of the pyramid is the $(n-1)$ dimensional object $B$ specified by $\left(\mathbf{e}_{1}+\mathbf{e}_{2}+\cdots+\mathbf{e}_{n}\right) \cdot \mathbf{x}=1, \mathbf{e}_{i} \cdot \mathbf{x} \geqslant 0$ for $i=1, \ldots, n$.

Find the point $C$ in $B$ equidistant from each vertex of $B$ and find the length of $O C .(C$ is the centroid of $B$.)

Show, using the Cauchy-Schwarz inequality, that this is the closest point in $B$ to the origin $O$.

Calculate the angle between $O C$ and any edge of the pyramid connected to $O$. What happens to this angle and to the length of $O C$ as $n$ tends to infinity?