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 $\| \vec{u} + \vec{v} \| ≤ \| \vec{u} \| + \| \vec{v} \|$.
(1)
\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:

Screen%20Shot%202013-12-11%20at%202.15.01%20AM.png

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 $\mid \vec{u} \cdot \vec{v} \mid ≤ \| \vec{u} \| \| \vec{v} \|$.
(2)
\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.
(3)
\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.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License