Page 14 - MATINF Nr. 7
P. 14

14                                                                                D. M˘arghidanu



            Definition 1 (synchrony/asynchrony). Two finite sequences (a 1 , a 2 , . . . , a n ) and (b 1 , b 2 , . . . , b n )
            of real numbers are called synchronous (or similarly ordered) if we have simultaneously:
            a 1 ≤ a 2 ≤ . . . ≤ a n and b 1 ≤ b 2 ≤ . . . ≤ b n , respectively a 1 ≥ a 2 ≥ . . . ≥ a n and b 1 ≥ b 2 ≥ . . . ≥
            b n (so if both sequences are simultaneously ascending or simultaneously descending, in other
            words if the two sequences have the same type of monotony). If the two sequences have inverse
            monotony we will call them asynchronous (or inversely ordered).


                Regarding the arithmetic mean, we can formulate the following result.

            Proposition 1. If a i ≥ 0, b i > 0 ∀i = 1, n are such that the sequences (a 1 , a 2 , . . . , a n ) and
            (b 1 , b 2 , . . . , b n ) are asynchronous, then
                                                                               ‹
                                       A n (a 1 , a 2 , . . . , a n )  a 1 a 2  a n
                                                          ≤ A n     ,  , . . . ,  .
                                       A n (b 1 , b 2 , . . . , b n )  b 1 b 2  b n


                The above inequality, even in a refined form (but apparently under slightly restricted
            conditions) was obtained by the author in [6], with the following statement:

            Proposition 2. If a i ≥ 0, b i > 0 satisfies the relationship a i + b i = C, ∀ i = 1, n, C ∈ R, then
                                                                          ‹                      ‹
                A n (a 1 , a 2 , . . . , a n )                1 1        1           a 1 a 2     a n
                                   ≤ A n (a 1 , a 2 , . . . , a n ) · A n  ,  , . . . ,  ≤ A n  ,  , . . . ,  .  (1)
                 A n (b 1 , b 2 , . . . , b n )               b 1 b 2    b n         b 1 b 2     b n
            Proof. The inequality from the left holds even without the condition a i +b i = C. Indeed, using
                                                                                                P 1
                                                                                                 n
                                                                                                         n
                                                                                                i=1  b i
            the inequality between the arithmetic mean and the harmonic mean, we have                ≥  n    .
                                                                                                  n     P
                                                                                                           b i
                                      ‹                                                               i=1
                         1 1         1              1
            Hence, A n     ,   , . . . ,  ≥                  . By multiplication with A n (a 1 , a 2 , . . . , a n ), the
                         b 1 b 2    b n     A n (b 1 , b 2 , . . . , b n )
            inequality on the left is obtained. Equality happens for b 1 = b 2 = . . . = b n .

                To establish the inequality on the right, we will use Cebyshev’s Inequality. Indeed, without loss
                                                                   ⇔                                      ⇔
            of generality, we can assume that a 1 ≤ a 2 ≤ . . . ≤ a n    C−a 1 ≥ C−a 2 ≥ . . . ≥ C−a n
                                                1     1            1
            ⇔     b 1 ≥ b 2 ≥ . . . ≥ b n  ⇔       ≤     ≤ . . . ≤   , so the sequences (a 1 , a 2 , . . . , a n ) and
                                                b 1   b 2          b n
                                                                            n      P 1      n      1
                                                                                   n
                           ‹                                              P   a i          P  a i ·
               1 1        1                                                i=1     i=1  b i  i=1   b i
                ,   , . . . ,  are synchronous. By Chebyshev’s Inequality        ·       ≤           , that is
              b 1 b 2    b n                     ‹                      ‹  n       n         n
                                     1 1        1           a 1 a 2     a n
            A n (a 1 , a 2 , . . . , a n ) · A n  ,  , . . . ,  ≤ A n  ,  , . . . ,  , which complete the proof.
                                     b 1 b 2    b n         b 1 b 2     b n
                A similar solution to the problem can be found in [6] (offered in the group of Nassim Nicholas
            Taleb).

            Remark 2. The condition a i + b i = C ∀ i = 1, n, C ∈ R, occurs only to establish that the
                                                            ‹
                                            
                                              1 1         1
            sequences (a 1 , a 2 , . . . , a n ) and  ,  , . . . ,  are similarly ordered. Obviously, it can be
                                              b 1 b 2    b n
            replaced from the beginning in the statement of Proposition 2 with the less restrictive condition
            that (a 1 , a 2 , . . . , a n ) and (b 1 , b 2 , . . . , b n ) to be inversely ordered. In this way it is possible to
            apply Chebyshev Inequality.

                We are now able to establish similar results for harmonic mean.
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