Question

Prove the triangle inequality \(\|\mathbf{u}+\mathbf{v}\| \leq\|\mathbf{u}\|+\|\mathbf{v}\|\).

Short answer

The triangle inequality \( \|\mathbf{u}+\mathbf{v}\| \leq \|\mathbf{u}\|+\|\mathbf{v}\| \) is proven by expanding the left side, using the dot product property and simplifying the expression.

Step-by-step solution

Expand Norm

Begin by expanding the left-hand side of the triangle inequality. Apply the definition of a norm, \( \|\mathbf{u}+\mathbf{v}\|^2 = (\mathbf{u}+\mathbf{v})\cdot(\mathbf{u}+\mathbf{v}) = \mathbf{u}\cdot\mathbf{u} + 2\mathbf{u}\cdot\mathbf{v} + \mathbf{v}\cdot\mathbf{v} \).

Use Dot Product Property

Use the dot product property \(\mathbf{u}\cdot\mathbf{v} \leq \|\mathbf{u}\|\|\mathbf{v}\|\). Substitute this into the expression to get \(\|\mathbf{u}\|^2 + 2\mathbf{u}\cdot\mathbf{v} + \|\mathbf{v}\|^2 \leq \|\mathbf{u}\|^2 + 2\|\mathbf{u}\|\|\mathbf{v}\| + \|\mathbf{v}\|^2\).

Simplify and Take the Square Root

Simplify the expression and take the square root on both sides. This gives \(\|\mathbf{u}+\mathbf{v}\| \leq \sqrt{\|\mathbf{u}\|^2 + 2\|\mathbf{u}\|\|\mathbf{v}\| + \|\mathbf{v}\|^2} = \|\mathbf{u}\| + \|\mathbf{v}\|\).