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?


  • 1
    $\begingroup$ Read it backwards, starting with the (presumably) well known $\;\;-1 \le \cos \theta \le 1$ $\;\;\iff\;\; \mid \cos \theta \mid \leq 1$ $\;\;\implies\;\; \mid \|\vec{x}\|\|\vec{y}\|\cos \theta \mid \leq \|\vec{x}\|\|\vec{y}\|$ $\;\;\iff\;\; \mid\vec{x}\cdot\vec{y}\mid \leq \|\vec{x}\|\|\vec{y}\|\,$. $\endgroup$
    – dxiv
    Jun 26, 2021 at 6:23
  • 2
    $\begingroup$ The 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. $\endgroup$
    – user169852
    Jun 26, 2021 at 6:36


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