Cauchy's Inequality/Vector Form

Theorem

Let $\mathbf a$ and $\mathbf b$ be vectors in a vector space $V$.

Then:

$\paren {\mathbf a \cdot \mathbf b}^2 \le \paren {\mathbf a \cdot \mathbf a} \paren {\mathbf b \cdot \mathbf b}$

where $\cdot$ denotes dot product.

Let us express $\mathbf a$ and $\mathbf b$ in component form:

$\ds \mathbf a$

$:=$

$\ds \sum_{i \mathop = 1}^n a_i \mathbf e_i$

$\ds \mathbf b$

$:=$

$\ds \sum_{i \mathop = 1}^n b_i \mathbf e_i$

where the ordered $n$-tuple $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is the standard ordered basis of $V$.

Then we have:

$\ds \paren {\mathbf a \cdot \mathbf b}^2$

$=$

$\ds \paren {\sum_{i \mathop = 1}^n a_i b_i}^2$

Definition of Dot Product

$\ds $

$\le$

$\ds \paren {\sum_{i \mathop = 1}^n {a_i}^2} \paren {\sum_{i \mathop = 1}^n {b_i}^2}$

$\ds $

$\le$

$\ds \paren {\mathbf a \cdot \mathbf a} \paren {\mathbf b \cdot \mathbf b}$

Definition of Dot Product

$\blacksquare$