Question

Prove that if $\mathbf{u}, \mathbf{v}$, and $\mathbf{w}$ are vectors in $R^{n}$, then: $(a) \quad\|\mathbf{u}+\mathbf{v}\| \leq\|\mathbf{u}\|+\|\mathbf{v}\| \quad$ [Triangle inequality for vectors] $(b) \quad d(\mathbf{u}, \mathbf{v}) \leq d(\mathbf{u}, \mathbf{w})+d(\mathbf{w}, \mathbf{v}) \quad$ [Triangle inequality for distances]

Answer 1

Proof

(a) $$ \begin{aligned} \|\mathbf{u}+\mathbf{v}\|^{2} &=(\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=(\mathbf{u} \cdot \mathbf{u})+2(\mathbf{u} \cdot \mathbf{v})+(\mathbf{v} \cdot \mathbf{v}) \\ &=\|\mathbf{u}\|^{2}+2(\mathbf{u} \cdot \mathbf{v})+\|\mathbf{v}\|^{2} \end{aligned} $$ $$ \leq\|\mathbf{u}\|^{2}+2|\mathbf{u} \cdot \mathbf{v}|+\|\mathbf{v}\|^{2} $$ $$ \begin{aligned} &\leq\|\mathbf{u}\|^{2}+2\|\mathbf{u}\|\|\mathbf{v}\|+\|\mathbf{v}\|^{2} \\ &=(\|\mathbf{u}\|+\|\mathbf{v}\|)^{2} \end{aligned} $$ This completes the proof since both sides of the inequality in part $(a)$ are nonnegative.

 

Proof

$(b)$ It follows from part (a) that $$ \begin{aligned} d(\mathbf{u}, \mathbf{v}) &=\|\mathbf{u}-\mathbf{v}\|=\|(\mathbf{u}-\mathbf{w})+(\mathbf{w}-\mathbf{v})\| \\ & \leq\|\mathbf{u}-\mathbf{w}\|+\|\mathbf{w}-\mathbf{v}\|=d(\mathbf{u}, \mathbf{w})+d(\mathbf{w}, \mathbf{v}) \end{aligned} $$