**Cauchy-Schwartz Inequality****Definition:**If \((a_1, a_2, \ldots, a_n)\) and \((b_1, b_2, \ldots, b_n)\) are sequences of real numbers, then the Cauchy-Schwartz inequality states that:\[\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right) \geq \left(\sum_{i=1}^n a_i b_i\right)^2\]with equality if and only if there exists some real number \(r\) such that \((a_1, a_2, \ldots, a_n) = r(b_1, b_2, \ldots, b_n)\).**Vector Notation:**If \(\mathbf{u} = (a_1, a_2, \ldots, a_n)\) and \(\mathbf{v} = (b_1, b_2, \ldots, b_n)\), then this inequality can also be expressed as:\[|\mathbf{u} \cdot \mathbf{v}| \leq \|\mathbf{u}\| \|\mathbf{v}\|\]**Proof Outline:**There are many ways to prove this inequality. 1. Note that:\[2\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right) - 2\left(\sum_{i=1}^n a_i b_i\right)^2 = \sum_{i=1}^n \sum_{j=1}^n (a_i b_j - a_j b_i)^2\]2. Recognize that:\[\sum_{i=1}^n \sum_{j=1}^n (a_i b_j - a_j b_i)^2 \geq 0\]This follows by the trivial inequality that squares are non-negative.3. By dividing by 2 and adding the second sum to both sides, we obtain:\[\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}

Name: **Cauchy-Schwartz Inequality****Definition:**If \((a_1, a_2, \ldots,
Uploaded: 2024-03-20T06:38:58.000Z
Channel: Bartleby
Description: **Cauchy-Schwartz Inequality****Definition:**If \((a_1, a_2, \ldots, a_n)\) and \((b_1, b_2, \ldots, b_n)\) are sequences of real numbers, then the Cauchy-Schwartz inequality states that:\[\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\righ