# Triangle Inequality/Vectors in Euclidean Space

## Theorem

Let $\mathbf x, \mathbf y$ be vectors in $\R^n$.

Let $\norm {\, \cdot \,}$ denote vector length.

Then:

$\norm {\mathbf x + \mathbf y} \le \norm {\mathbf x} + \norm {\mathbf y}$

If the two vectors are scalar multiples where said scalar is non-negative, an equality holds:

$\exists \lambda \in \R, \lambda \ge 0: \mathbf x = \lambda \mathbf y \iff \norm {\mathbf x + \mathbf y} = \norm {\mathbf x} + \norm {\mathbf y}$

## Proof

Let $\mathbf x, \mathbf y \in \R^n$.

We have:

 $\ds \norm {\mathbf x + \mathbf y}^2$ $=$ $\ds \paren {\mathbf x + \mathbf y} \cdot \paren {\mathbf x + \mathbf y}$ Dot Product of Vector with Itself $\ds$ $=$ $\ds \mathbf x \cdot \mathbf x + \mathbf x \cdot \mathbf y + \mathbf y \cdot \mathbf x + \mathbf y \cdot \mathbf y$ Dot Product Distributes over Addition $\ds$ $=$ $\ds \mathbf x \cdot \mathbf x + 2 \paren {\mathbf x \cdot \mathbf y} + \mathbf y \cdot \mathbf y$ Dot Product Operator is Commutative $\ds$ $=$ $\ds \norm {\mathbf x}^2 + 2 \paren {\mathbf x \cdot \mathbf y} + \norm {\mathbf y}^2$ Dot Product of Vector with Itself
 $\ds \size {\mathbf x \cdot \mathbf y}$ $\le$ $\ds \norm {\mathbf x} \norm {\mathbf y}$ $\ds \leadsto \ \$ $\ds \mathbf x \cdot \mathbf y$ $\le$ $\ds \norm {\mathbf x} \norm {\mathbf y}$ Negative of Absolute Value $\ds \leadsto \ \$ $\ds \norm {\mathbf x}^2 + 2 \paren {\mathbf x \cdot \mathbf y} + \norm {\mathbf y}^2$ $\le$ $\ds \norm {\mathbf x}^2 + 2 \paren {\norm {\mathbf x} \norm {\mathbf y} } + \norm {\mathbf y}^2$ multiply both sides with $2$, and add $\norm {\mathbf x}^2 + \norm {\mathbf y}^2$ to both sides $\ds$ $=$ $\ds \paren {\norm {\mathbf x} + \norm {\mathbf y} }^2$ $\ds \leadsto \ \$ $\ds \norm {\mathbf x + \mathbf y}^2$ $\le$ $\ds \paren {\norm {\mathbf x} + \norm {\mathbf y} }^2$ $\ds \leadsto \ \$ $\ds \norm {\mathbf x + \mathbf y}$ $\le$ $\ds \norm {\mathbf x} + \norm {\mathbf y}$ taking the square root of both sides

$\blacksquare$

To prove that the equality holds if the vectors are scalar multiples of each other, assume:

$\exists \lambda \in \R, \lambda \ge 0: \mathbf v = \lambda \mathbf w$

### Sufficient Condition

 $\ds \norm {\mathbf v + \mathbf w}$ $=$ $\ds \norm {\lambda \mathbf w + \mathbf w}$ $\ds$ $=$ $\ds \norm {\paren {\lambda + 1} \mathbf w}$ $\ds$ $=$ $\ds \paren {\lambda + 1} \norm {\mathbf w}$ $\ds$ $=$ $\ds \lambda \norm {\mathbf w} + 1 \norm {\mathbf w}$ $\ds$ $=$ $\ds \norm {\lambda \mathbf w} + \norm {1 \mathbf w}$ $\ds$ $=$ $\ds \norm {\mathbf v} + \norm {\mathbf w}$

$\Box$