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Definition (Similarly and oppositely ordered sequences)
Two sequences of real numbers (a1, . . . , an) and (b1, . . . , bn) are similarly ordered if and only if for each
pair (i, j), where 1 ≤ i, j ≤ n, we have
(ai − aj)(bi − bj) 0 (1)
in other words
(ai aj ∧ bi bj) ∨ (ai aj ∧ bi bj)
Analogically, the aforementioned sequences are oppositely ordered if and only if the inequality (1) is reversed.
For the purpose of the theorem formulated below, let’s introduce the following notation
n
k=1
akbk = a1b1 + a2b2 + · · · + anbn =
a1 a2 · · · an
b1 b2 · · · bn
Theorem (Inequalities between sums of the products of the sequences elements) :
If the sequences of real numbers (a1, . . . , an) and (b1, . . . , bn) are similarly ordered then for
any permutation (b1, . . . , bn) of the given sequence (b1, . . . , bn) we have
n
k=1
akbk =
a1 a2 · · · an
b1 b2 · · · bn
a1 a2 · · · an
b1 b2 · · · bn
=
n
k=1
akbk (2)
If the sequences of real numbers (a1, . . . , an) and (b1, . . . , bn) are oppositely ordered then for
any permutation (b1, . . . , bn) of the given sequence (b1, . . . , bn) we get the reversed version of
the inequality (2).
c 2015/10/13 22:41:36, Mikołaj Hajduk 1 / 1

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Permutation theorem and its use to proving inequalities.

  • 1. Definition (Similarly and oppositely ordered sequences) Two sequences of real numbers (a1, . . . , an) and (b1, . . . , bn) are similarly ordered if and only if for each pair (i, j), where 1 ≤ i, j ≤ n, we have (ai − aj)(bi − bj) 0 (1) in other words (ai aj ∧ bi bj) ∨ (ai aj ∧ bi bj) Analogically, the aforementioned sequences are oppositely ordered if and only if the inequality (1) is reversed. For the purpose of the theorem formulated below, let’s introduce the following notation n k=1 akbk = a1b1 + a2b2 + · · · + anbn = a1 a2 · · · an b1 b2 · · · bn Theorem (Inequalities between sums of the products of the sequences elements) : If the sequences of real numbers (a1, . . . , an) and (b1, . . . , bn) are similarly ordered then for any permutation (b1, . . . , bn) of the given sequence (b1, . . . , bn) we have n k=1 akbk = a1 a2 · · · an b1 b2 · · · bn a1 a2 · · · an b1 b2 · · · bn = n k=1 akbk (2) If the sequences of real numbers (a1, . . . , an) and (b1, . . . , bn) are oppositely ordered then for any permutation (b1, . . . , bn) of the given sequence (b1, . . . , bn) we get the reversed version of the inequality (2). c 2015/10/13 22:41:36, Mikołaj Hajduk 1 / 1