By S.S.Dragomir

**Read or Download A Survey on Cauchy-Buniakowsky-Schwartz Type Discrete Inequalities PDF**

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**Extra info for A Survey on Cauchy-Buniakowsky-Schwartz Type Discrete Inequalities**

**Example text**

N)and b be two sequences of complex numbers. Then one has the inequality: n n 2 |ai | i=1 |bi | − i=1 2 n 2 a ¯ i bi i=1 n ≥ i=1 |ai | |bi | bi − ai n n |ai | bi i=1 a ¯i bi ≥ 0. 21) i=1 44 CHAPTER 2. 10]). ¯ = (b1 , . . , bn ) be sequences of real Theorem 73 Let ¯ a = (a1 , . . , an ), b numbers and p ¯ = (p1 , . . , pn ), q ¯ = (q1 , . . , qn ) be sequences of nonnegative real numbers such that pk ≥ qk for any k ∈ {1, . . , n} . 22) k=1 n n qk a2k (pk − qk ) ak bk + ≥ k=1 k=1 n ≥ qk b2k (pk − qk ) ak bk + 2 k=1 2 n qk ak bk k=1 k=1 2 n ≥ 1 2 n p k ak b k .

A SECOND REFINEMENT IN TERMS OF MODULI n 2 i=1 ei condition holds n 41 = 1. 16) i=1 n n n ak¯bk − ≥ k=1 k=1 2 n ek¯bk + ak e¯k · k=1 n ek¯bk ak e¯k · k=1 k=1 2 n ak¯bk . ≥ k=1 The proof is similar to the one in Theorem 65 on using the corresponding (CBS) −inequality for complex numbers. Remark 68 Similar particular inequalities may be stated, but we omit the details. 3 A Second Refinement in Terms of Moduli The following lemma holds. Lemma 69 Let ¯ a = (a1 , . . , an ) be a sequence of real numbers and p ¯ = n (p1 , .

B1 , . . , bn ) be sequences of comTheorem 63 Let ¯ a = (a1 , . . , an ) and b plex numbers. Then one has the inequality n n 2 |ai | i=1 2 n 2 |bi | − i=1 ai b i i=1 n ≥ n n |ai | a ¯i i=1 n |bi | bi − i=1 |ai | bi i=1 |bi | a ¯i ≥ 0. 8) i=1 Proof. We have for any i, j ∈ {1, . . , n} that |¯ ai b j − a ¯j bi | ≥ ||ai | |bj | − |aj | |bi || . Multiplying by |¯ ai b j − a ¯j bi | ≥ 0, we get |¯ ai b j − a ¯j bi |2 ≥ ||ai | a ¯i |bj | bj + |aj | a ¯j |bi | bi − |ai | bi |bj | a ¯j − |bi | a ¯i |aj | bj | .