Validity of proof of Cauchy-Schwarz inequality

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SUMMARY

The proof of the Cauchy-Schwarz inequality presented in the discussion is valid under the condition that the relationship \( x \cdot y = |x| |y| \cos \theta \) is established. The proof correctly identifies that if either \( x \) or \( y \) is zero, the inequality holds trivially. For non-zero values, it effectively uses the properties of cosine to demonstrate that \( |x \cdot y| \leq |x| |y| \) is satisfied, provided the cosine definition is acknowledged or previously proven.

PREREQUISITES
  • Understanding of the Cauchy-Schwarz inequality
  • Familiarity with trigonometric identities, specifically the cosine function
  • Basic knowledge of vector operations and dot products
  • Experience with mathematical proofs and logical reasoning
NEXT STEPS
  • Study the formal proof of the Cauchy-Schwarz inequality in linear algebra
  • Explore the implications of the inequality in various mathematical contexts
  • Learn about the geometric interpretation of the dot product and cosine
  • Review related inequalities such as the triangle inequality and their proofs
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Mathematicians, students studying linear algebra, and anyone interested in understanding the foundations of inequalities in mathematics.

HaniZaheer
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Proof: If either x or y is zero, then the inequality |x · y| ≤ | x | | y | is trivially correct because both sides are zero.
If neither x nor y is zero, then by x · y = | x | | y | cos θ,
|x · y|=| x | | y | cos θ | ≤ | x | | y |
since -1 ≤ cos θ ≤ 1

How valid is this a proof of the Cauchy-Schwarz inequality?
 
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It looks good as long as x · y = | x | | y | cos θ is given by definition or is already proven. (It's good form to indicate "by definition" or "by Lemma xxx", etc.)
 
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Alright, thanks a lot
 

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