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# Use Theorem 3 to prove the Cauchy-Schwarz Inequality:$$\mid a \cdot b \mid \le \mid a \mid \mid b \mid$$

## $$\mathbf{a} \cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}| \cos (\theta)$$$$|\mathbf{a} \cdot \mathbf{b}|=|\mathbf{a}||\mathbf{b}||\cos (\theta)|$$And, since $|\cos (\theta)| \leq 1,$ thus we have$$\leq|\mathbf{a}||\mathbf{b}| \quad (\mathrm{R} . \mathrm{T} . \mathrm{P})$$

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YZ

Yushan Z.

September 26, 2021

Thanks!

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### Video Transcript

Yeah. In this collision we are going to grow The 40 Schwarz inequality or no, eight B Less than equals two more or mm. No, 1st. We consider that towards product. So now on applying uh, Children on gold Rolex eager. Don't be a fool student. More A or B was. Yeah. Does uh, ford product of more A dod E is equals two. More morty would be was Cheeto? Um, We continue you to about the staff as more clothes. Kita is less than equals to one, then hold a door. Yeah. Is first an equals true or or B are by Why? No, from Hubble. It is proved that or or E less than equal Stream more A more mm.

Dr. A.P.J Abdul Kalam Technical University

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