# 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:

 $\displaystyle \norm {\mathbf x + \mathbf y}^2$ $=$ $\displaystyle \paren {\mathbf x + \mathbf y} \cdot \paren {\mathbf x + \mathbf y}$ Dot Product of Vector with Itself $\displaystyle$ $=$ $\displaystyle \mathbf x \cdot \mathbf x + \mathbf x \cdot \mathbf y + \mathbf y \cdot \mathbf x + \mathbf y \cdot \mathbf y$ Dot Product Distributes over Addition $\displaystyle$ $=$ $\displaystyle \mathbf x \cdot \mathbf x + 2 \paren {\mathbf x \cdot \mathbf y} + \mathbf y \cdot \mathbf y$ Dot Product Operator is Commutative $\displaystyle$ $=$ $\displaystyle \norm {\mathbf x}^2 + 2 \paren {\mathbf x \cdot \mathbf y} + \norm {\mathbf y}^2$ Dot Product of Vector with Itself
 $\displaystyle \size {\mathbf x \cdot \mathbf y}$ $\le$ $\displaystyle \norm {\mathbf x} \norm {\mathbf y}$ $\displaystyle \leadsto \ \$ $\displaystyle \mathbf x \cdot \mathbf y$ $\le$ $\displaystyle \norm {\mathbf x} \norm {\mathbf y}$ Negative of Absolute Value $\displaystyle \norm {\mathbf x}^2 + 2 \paren {\mathbf x \cdot \mathbf y} + \norm {\mathbf y}^2$ $\le$ $\displaystyle \norm {\mathbf x}^2 + 2 \paren {\norm {\mathbf x} \norm {\mathbf y} } + \norm {\mathbf y}^2$ $\displaystyle$ $=$ $\displaystyle \paren {\norm {\mathbf x} + \norm {\mathbf y} }^2$ $\displaystyle \leadsto \ \$ $\displaystyle \norm {\mathbf x + \mathbf y}^2$ $\le$ $\displaystyle \paren {\norm {\mathbf x} + \norm {\mathbf y} }^2$ $\displaystyle \leadsto \ \$ $\displaystyle \norm {\mathbf x + \mathbf y}$ $\le$ $\displaystyle \norm {\mathbf x} + \norm {\mathbf y}$ 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

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

$\Box$