I found this proof for Cauchy-Schwarz on this forum. Can someone help explain how does the absolute value of |cos($\theta$) < 1| help prove the inequality?
Proofs of the Cauchy-Schwarz Inequality?
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1Read it backwards, starting with the (presumably) well known $;;-1 \le \cos \theta \le 1$ $;;\iff;; \mid \cos \theta \mid \leq 1$ $;;\implies;; \mid |\vec{x}||\vec{y}|\cos \theta \mid \leq |\vec{x}||\vec{y}|$ $;;\iff;; \mid\vec{x}\cdot\vec{y}\mid \leq |\vec{x}||\vec{y}|,$. – dxiv Jun 26 '21 at 06:23
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2The second line is wrong. $|x \cdot y| = |x| |y| |\cos \theta|$, not $|x| |y| \cos \theta$. More importantly, this argument seems potentially circular, depending on how $\theta$ is defined. – Jun 26 '21 at 06:36