Triangle Inequality for Vector Norms¶

Theorem¶

For vectors $\boldsymbol{x},\boldsymbol{y}\in\mathbb{R}^n$${\mathbb{R}}^n$, the triangle inequality states $\|\boldsymbol{x}+\boldsymbol{y}\|\leq\|\boldsymbol{x}\|+\|\boldsymbol{y}\|$$\Vert \mathbf{y}\Vert$.

Interpretation¶

The triangle inequality says that the direct route is never longer than a detour: the length of one side of a triangle is at most the sum of the lengths of the other two sides.

In the upper triangle, $\boldsymbol{y}$ starts at the tip of $\boldsymbol{x}$, so the third side is $\boldsymbol{x}+\boldsymbol{y}$$\mathbf{y}$, and its length $\|\boldsymbol{x}+\boldsymbol{y}\|$ is strictly smaller than $\|\boldsymbol{x}\|+\|\boldsymbol{y}\|$$\Vert \mathbf{y}\Vert$. In the flattened triangle below, $\boldsymbol{y}=c\boldsymbol{x}$ with $c\ge0$ points the same way as $\boldsymbol{x}$, and the lengths add exactly. For nonzero vectors this is the only case of equality; equality also holds when either vector is the zero vector.

The inequality follows from the Cauchy-Schwarz inequality. Expanding with the inner product gives $\|\boldsymbol{x}+\boldsymbol{y}\|^2=\|\boldsymbol{x}\|^2+2\,\boldsymbol{x}\cdot\boldsymbol{y}+\|\boldsymbol{y}\|^2$$\Vert \mathbf{y}{\Vert }^2$, and Cauchy-Schwarz bounds the middle term by $2\|\boldsymbol{x}\|\|\boldsymbol{y}\|$, so the right-hand side is at most $(\|\boldsymbol{x}\|+\|\boldsymbol{y}\|)^2$$\Vert \mathbf{y}\Vert {)}^2$. Equality needs $\boldsymbol{x}\cdot\boldsymbol{y}=\|\boldsymbol{x}\|\|\boldsymbol{y}\|$$\Vert \mathbf{x}\Vert \Vert \mathbf{y}\Vert$, an angle of $0$ between nonzero vectors.

For $n=1$ the Euclidean norm is the absolute value, and the statement is the triangle inequality $|a+b|\le|a|+|b|$ for numbers. Applied to $\boldsymbol{x}-\boldsymbol{z}=(\boldsymbol{x}-\boldsymbol{y})+(\boldsymbol{y}-\boldsymbol{z})$$\mathbf{z})$, it says that for the Euclidean distance a stop at $\boldsymbol{y}$ never shortens the trip: $d(\boldsymbol{x},\boldsymbol{z})\le d(\boldsymbol{x},\boldsymbol{y})+d(\boldsymbol{y},\boldsymbol{z})$$d(\mathbf{y},\mathbf{z})$. Later it becomes one of the rules that every norm must satisfy.

The inequality is only an upper bound: for $(1,0)$ and $(-1,0)$ the sum is $\boldsymbol{0}$, with norm $0$, not $2$.

Key Results¶

• The triangle inequality follows from the Cauchy-Schwarz inequality.
• It is one of the defining properties of a norm.

Examples¶

• $\|(1,0)+(0,1)\|=\sqrt{2}\leq2$$2$.
• If $\boldsymbol{x}$ and $\boldsymbol{y}$ point in the same direction, equality can hold.

Prerequisites¶

• Euclidean Norm
• Cauchy-Schwarz Inequality

Related Concepts¶

• Euclidean Distance
• Triangle Inequality

Sources¶

• Math II: Linear Algebra · Week 5

Keywords: Minkowski inequality Symbols: element of, real numbers, square root

Last updated: 20 October 2026 03:02