Triangle Inequality

There are two forms of the triangle inequality, one in terms of lengths (norms) and one in terms of distances. We will look at both and prove them.

# Triangle Inequality (Lengths)

Theorem 1 (Triangle Inequality for Vector Lengths): For any two vectors $\vec{u}, \vec{v} \in \mathbb{R}^n$ such that $\vec{u} \perp \vec{v}$, it follows that $\| \vec{u} + \vec{v} \| ≤ \| \vec{u} \| + \| \vec{v} \|$. |

\begin{align} ||\vec{u} + \vec{v} || \le ||\vec{u} || + || \vec{v} || \end{align}

Imagine that we have a triangle constructed from vectors. We will omit the coordinate axes for better visualization:

It's rather clear that the sum of the lengths of **a** and **b** is greater than that of the length of the vector $(\vec{a} + \vec{b})$ (the longest side of the triangle).

**Proof:**This proof will be purely algebraic and uses the Cauchy-Schwarz Inequality that says $\mid \vec{u} \cdot \vec{v} \mid ≤ \| \vec{u} \| \| \vec{v} \|$.

\begin{align} \| \vec{u} + \vec{v} \|^2 = (\vec{u} + \vec{v}) \cdot (\vec{u} + \vec{v}) \\ \| \vec{u} + \vec{v} \|^2 = \| \vec{u} \|^2 +2(\vec{u} \cdot \vec{v}) + \| \vec{v} \|^2 \\ \| \vec{u} + \vec{v} \|^2 ≤ \|\vec{u}\|^2 + 2\mid \vec{u} \cdot \vec{v}\mid + \|\vec{v}||^2 \\ \| \vec{u} + \vec{v} \|^2 ≤ \|\vec{u}\|^2 + 2\| \vec{u} \| \| \vec{v} \| + \|\vec{v}\|^2 \\ \| \vec{u} + \vec{v} \|^2 ≤ (\| \vec{u} \| + \| \vec{v} \|)^2 \\ \| \vec{u} + \vec{v} \| ≤ \| \vec{u} \| + \| \vec{v} \| \\ \blacksquare \end{align}

# Triangle Inequality (Distances)

Recall that $d(\vec{u}, \vec{v})$ represents the distance between the points represented by the vectors $\vec{u}$ and $\vec{v}$.

Theorem 2 (Triangle Inequality for Distances): For any vectors $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^n$, the distance $d(\vec{u}, \vec{v})$ is the shortest distance from $\vec{u}$ to $\vec{v}$, that is $d(\vec{u}, \vec{v}) ≤ d(\vec{u}, \vec{w}) + d(\vec{w}, \vec{v})$. |

**Proof:**Recall that $d(\vec{u}, \vec{v}) = \| \vec{u} - \vec{v} \|$. The rest of the proof is rather straightforward.

\begin{align} d(\vec{u}, \vec{v}) = \| \vec{u} - \vec{v} \| \\ d(\vec{u}, \vec{v}) = \| (\vec{u} - \vec{w}) + (\vec{w} - \vec{v}) \| \\ d(\vec{u}, \vec{v}) ≤ \| \vec{u} - \vec{w} \| + \| \vec{w} - \vec{v} \| \\ d(\vec{u}, \vec{v}) ≤ d(\vec{u}, \vec{w}) + d(\vec{w}, \vec{v}) \\ \blacksquare \end{align}

(4)
\begin{align} || \vec{u} + \vec{v} ||^2 \leq ||\vec{u}||^2 + 2|\vec{u} \cdot \vec{v}| + ||\vec{v}||^2 \end{align}

With this proof we note that the shortest distance between two points is always the direct line connecting them.