Chapter 11: Problem 87
Cauchy-Schwarz Inequality The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\) (because \(|\cos \theta| \leq 1\) ). This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Triangle Inequality Consider the vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{u}+\mathbf{v}\) (in any number of dimensions). Use the following steps to prove that \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}|\) a. Show that \(|\mathbf{u}+\mathbf{v}|^{2}=(\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+\) \(2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2}\) b. Use the Cauchy-Schwarz Inequality to show that \(|\mathbf{u}+\mathbf{v}|^{2} \leq(|\mathbf{u}|+|\mathbf{v}|)^{2}\) c. Conclude that \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}|\) d. Interpret the Triangle Inequality geometrically in \(\mathbb{R}^{2}\) or \(\mathbb{R}^{3}\).
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.