# Statement(s) > [!Lemma] Theorem (Euclidean Norm Satisfies Triangle Inequality) > Let $n\geq 1.$ Let $\underline{v},\underline{w} \in \mathbb{R}^{n}$ be [[Real n-Space|real vectors]],then their [[Euclidean Norm|length]] satisfies $||\underline{v}+\underline{w}|| \leq ||v||+||w||\tag{1}$and equality holds if $\underline{v}$ and $\underline{w}$ are [[Collinearity of Two Vectors|collinear]]. > [!NOTE] Theorem (Euclidean metric satisfies triangle inequality) > $(1)$ implies $\lVert x-y \rVert \leq \lVert x-z \rVert + \lVert y-z \rVert $and vice versa using [[Euclidean Metric is Translation Invariant|translation of the Euclidean metric]] so the Euclidean metric satisfies the triangle inequality # Proof(s) ###### Proof of statement 1: We compute the length-squared of $\underline{v}+\underline{w}:$ $\begin{align}||\underline{v}+\underline{w}||^{2} &= (\underline{v}+\underline{w}) \cdot (\underline{v}+\underline{w}) \\ &= \underline{v} \cdot \underline{v} + \underline{v} \cdot \underline{w} + \underline{w}\cdot\underline{v} + \underline{w} \cdot \underline{w} \\ & \leq ||\underline{v}||^{2} + 2 || \underline{v}|| \,||w || + ||\underline{w} ||^{2} \tag{*} \\ &= (||\underline{v}||+||\underline{w}||)^{2} \end{align}.$Note that $(*)$ uses [[Upper Bound for Absolute Value of Dot Product in Real n-Space (Cauchy-Schwartz Inequality)]]. Taking the positive square root gives us the result.