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Numerade Educator

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Problem 61 Medium Difficulty

Use Theorem 3 to prove the Cauchy-Schwarz Inequality:

$$ \mid a \cdot b \mid \le \mid a \mid \mid b \mid $$

Answer

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

Discussion

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YZ

Yushan Z.

September 26, 2021

Thanks!

Video Transcript

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