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proof of Cauchy-Schwarz inequality for real numbers
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(Proof)
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The version of the Cauchy-Schwartz inequality we want to prove is$$ \left(\sum_{k=1}^n a_k b_k \right)^2 \le \sum_{k=1}^n a_k^2 \cdot \sum_{k=1}^n b_k^2,$$ where the $a_k$ and $b_k$ are real numbers, with equality holding only in the case of proportionality, $a_k=\lambda b_k$ for some real $\lambda$ for all $k$ .
The proof is by direct calculation:
The above identity implies that the Cauchy-Schwarz inequality holds. Moreover, it is an equality only when$$ a_k b_l - a_l b_k = 0 \quad \Longleftrightarrow \quad \frac{a_k}{b_k} = \frac{a_l}{b_l} \text{ or } \frac{b_k}{a_k} = \frac{b_l}{a_l} \text{ or } a_k = b_k = 0,$$ for all $k$ and $l$ . In other words, equality holds only when $a_k = \lambda b_k$ for all $k$ for some real number $\lambda$ .
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Cross-references: Cauchy-Schwarz inequality, implies, identity, proof, equality, real numbers, Cauchy-Schwartz inequality
This is version 2 of proof of Cauchy-Schwarz inequality for real numbers, born on 2005-01-11, modified 2005-02-10.
Object id is 6635, canonical name is ProofOfCauchySchwarzInequalityForRealNumbers.
Accessed 8942 times total.
Classification:
| AMS MSC: | 15A63 (Linear and multilinear algebra; matrix theory :: Quadratic and bilinear forms, inner products) |
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Pending Errata and Addenda
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