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23) that  1 2 n 1 2 n pk a2k  pk b2k k=1 1 2 n qk a2k − k=1 1 2 n qk b2k k=1 2  k=1 2 n ≥ (pk − qk ) ak bk . 22). The other inequalities are obvious and we omit the details. 11]. CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY 46 ¯ be sequences of real numbers and ¯ Corollary 74 Let ¯ a, b s = (s1 , . . , sn ) be such that 0 ≤ sk ≤ 1 for any k ∈ {1, . . , n} . 25)  k=1 2 (1 − sk ) ak bk + k=1 s k ak b k k=1 2 n ≥ sk b2k n ≥ 1 2 n n b2k ≥  a2k k=1 n ak b k . k=1 ¯ and ¯ Remark 75 Assume that ¯ a, b s are as in Corollary 74.

N} . 6 A Refinement for Non-Constant Sequences The following result was proved in [3, Theorem 1]. 6. A REFINEMENT FOR NON-CONSTANT SEQUENCES 51 (i) ai = aj and bi = bj for i = j, i, j ∈ N; (ii) pi > 0 for all i ∈ N. 45) 2 pj . Proof. We shall follow the proof in [3]. Let J be a part of H. Define the mapping fJ : R → R given by  pi a2i  fJ (t) = i∈H i∈H\J  pi (bi + t)2  pi b2i + i∈J 2  − i∈H\J p i ai b i + pi ai (bi + t) . i∈J Then by the (CBS) −inequality we have that fJ (t) ≥ 0 for all t ∈ R.

REFINEMENTS OF THE (CBS) −INEQUALITY Proof. We will follow the proof in [2]. 32). 6]. ¯ = (b1 , . . , bn ) and ¯ Theorem 77 Let ¯ a = (a1 , . . , an ), b c = (c1 , . . , cn ) be sequences of real numbers such that (i) |bk | + |ck | = 0 (k ∈ {1, . . , n}) (ii) |ak | ≤ 2|bk ck | |bk |+|ck | for any k ∈ {1, . . , n} . Then one has the inequality n |ak | ≤ 2 k=1 n k=1 |bk | n k=1 (|bk | n k=1 |ck | . 33) Proof. We will follow the proof in [2]. By (ii) we observe that   2 |bk | 2 |bk ck | for any k ∈ {1, .

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