Cauchy-Schwarz Inequality

# Cauchy-Schwarz Inequality

We are now going to look at a very important inequality known as the **Cauchy-Schwarz Inequality**.

Theorem 1 (The Cauchy-Schwarz Inequality): For any two vectors $\vec{u}, \vec{v} \in \mathbb{R}^n$ then $\| \vec{u} \cdot \vec{v} \| ≤ \| \vec{u} \| \| \vec{v} \|$. |

**Proof:**We know from the dot product that $\vec{u} \cdot \vec{v} = \| \vec{u} \| \| \vec{v} \| \cos \theta$. If we take the absolute value of both sides of this formula we get:

\begin{align} \mid \vec{u} \cdot \vec{v} \mid = \mid \| \vec{u} \| \| \vec{v} \| \cos \theta \mid \\ \mid \vec{u} \cdot \vec{v} \mid = \| \vec{u} \| \| \vec{v} \| \mid \cos \theta \mid \\ \end{align}

- We note that $0 ≤ \: \mid \cos \theta \mid \: ≤ 1$. Thus the product on the righthand side will always be less or equal to $\| \vec{u} \| \| \vec{v} \|$, thus $\| \vec{u} \| \| \vec{v} \| \mid \cos \theta \mid ≤ \| \vec{u} \| \| \vec{v} \|$, or as the Cauchy-Schwarz inequality states: $\| \vec{u} \cdot \vec{v} \| ≤ \| \vec{u} \| \| \vec{v} \|$. $\blacksquare$