Use Theorem 3 to prove the Cauchy-Schwarz Inequality: $$ \mid a \cdot b \mid \le \mid a \mid \mid b \mid $$
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Let \( a \) and \( b \) be vectors in \( \mathbb{R}^n \). The dot product of \( a \) and \( b \) is defined as \( a \cdot b = a_1b_1 + a_2b_2 + \cdots + a_nb_n \), where \( a_i \) and \( b_i \) are the components of vectors \( a \) and \( b \), respectively. Show more…
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$$ \begin{array}{c}{\text { Use Theorem 3 to prove the Cauchy-Schwarz Inequality: }} \\ {|\mathbf{a} \cdot \mathbf{b}| \leqslant|\mathbf{a}||\mathbf{b}|}\end{array} $$
Use Theorem 3 to prove the Cauchy-Schwarz Inequality: $$|\mathbf{a} \cdot \mathbf{b}| \leqslant|\mathbf{a}||\mathbf{b}|$$
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