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Cauchy-Schwartz inequality • If (a1, a2, ..., an) and (b1, b2, ..., bn) are sequences of real numbers, then (E 7)(E b}) > (E-1 a;b;)² • with equality if and only if there exists some real number r such that (a1, a2, . .., an) = r(bị , b2, ... , bn) • If u = (a1, a2, ..., an) and v = (b1, b2, ..., bn), then this inequality says that |u · v| < || u||| v|| There are many ways to prove this > First note that 2(Σ 1f)(ΣΗ) -2(ΣΗabi) E Ei(a;b; – a;b;)² i=1 |

Cauchy-Schwartz inequality • If (a1, a2, ..., an) and (b1, b2, ..., bn) are sequences of real numbers, then (E 7)(E b}) > (E-1 a;b;)² • with equality if and only if there exists some real number r such that (a1, a2, . .., an) = r(bị , b2, ... , bn) • If u = (a1, a2, ..., an) and v = (b1, b2, ..., bn), then this inequality says that |u · v| < || u||| v|| There are many ways to prove this > First note that 2(Σ 1f)(ΣΗ) -2(ΣΗabi) E Ei(a;b; – a;b;)² i=1 |

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.6: Inequalities
Problem 80E
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Cauchy-Schwartz inequality • If (a1, a2,..., an) and (b1, b2, ..., bn) are sequences of real numbers, then (ΣΗ ΣΗ ) (ΣΗb)? • with equality if and only if there exists some real number r such that (a1, a2,..., an) = r(b1, b2, bn) • If u = (a1, a2,... , an) and v = (b1, b2, ..., bn), then this inequality says that |u · v < ||u|| || ... . • There are many ways to prove this > First note that 2(Σ1 )Σ - 2Σb)? - EL EL(aib; – a; b;)? • E-1 E-1(a¡b; – ajb;)² > 0 by the trivial inequality • Combining, dividing by 2, and adding the second sum to both sides gives (ΣΗ Σ ι bf) (Σ1 θ; b)? • Problem: Why do we get the part about the equality case? -
Transcribed Image Text:Cauchy-Schwartz inequality • If (a1, a2,..., an) and (b1, b2, ..., bn) are sequences of real numbers, then (ΣΗ ΣΗ ) (ΣΗb)? • with equality if and only if there exists some real number r such that (a1, a2,..., an) = r(b1, b2, bn) • If u = (a1, a2,... , an) and v = (b1, b2, ..., bn), then this inequality says that |u · v < ||u|| || ... . • There are many ways to prove this > First note that 2(Σ1 )Σ - 2Σb)? - EL EL(aib; – a; b;)? • E-1 E-1(a¡b; – ajb;)² > 0 by the trivial inequality • Combining, dividing by 2, and adding the second sum to both sides gives (ΣΗ Σ ι bf) (Σ1 θ; b)? • Problem: Why do we get the part about the equality case? -
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