Triangle Inequality/Vectors in Euclidean Space

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Theorem

Let $\mathbf x, \mathbf y$ be vectors in the real Euclidean space $\R^n$.

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

From the Cauchy-Bunyakovsky-Schwarz Inequality:

$\ds \size {\mathbf x \cdot \mathbf y}$

$\le$

$\ds \norm {\mathbf x} \norm {\mathbf y}$

$\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 $

$=$

$\ds \norm {\mathbf v} + \norm {\mathbf w}$

$\Box$

Necessary Condition

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Sources

2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): triangle inequality (for vectors)
2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): triangle inequality

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