Page 15 - MATINF Nr. 7
P. 15

Means of ratios versus ratios of means                                                         15



            Proposition 3. 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
                                                                               ‹
                                       H n (b 1 , b 2 , . . . , b n )  a 1 a 2  a n
                                                          ≥ H n     ,   , . . . ,  .                      (2)
                                       H n (a 1 , a 2 , . . . , a n )  b 1 b 2  b n

                The proof results here too from the refined form of this inequality captured in the following.


            Proposition 4. 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
                                                        ‹                                         ‹
                H n (b 1 , b 2 , . . . , b n )  1  1   1                              a 1 a 2    a n
                                   ≥ H n     ,   , . . . ,  · H n (b 1 , b 2 , . . . , b n ) ≥ H n  ,  , . . . ,  .  (3)
                H n (a 1 , a 2 , . . . , a n )  a 1 a 2  a n                          b 1 b 2    b n
                                                 1        1
            Proof. With the substitutions a i →    , b i →  , ∀ i = 1, n, (substitutions that keep the reverse -
                                                 b i     a i
            ordering of sequences), we have the following transformations:

                                                           n
                                                          P 1
                                    A n (a 1 , a 2 , . . . , a n ) →  i=1  b i  =  1  ,                   (4)
                                                            n      H n (b 1 , b 2 , . . . , b n )

                                                          n
                                                         P 1
                                                                           1
                                                            a i
                                                         i=1
                                    A n (b 1 , b 2 , . . . , b n ) →  =             ,                     (5)
                                                           n      H n (a 1 , a 2 , . . . , a n )
                                                               n
                                                     n
                                                    P 1       P
                                              ‹                 a i
                                  1 1        1      i=1  b i  i=1               1
                             A n    ,  , . . . ,  =       →         =       1  1       1  ‹,             (6)
                                  b 1 b 2    b n      n         n
                                                                      H n     ,   , . . . ,
                                                                            a 1 a 2    a n
                                           ‹                       ‹
                              a 1 a 2     a n          b 1 b 2     b n              1
                         A n     ,  , . . . ,  → A n     ,   , . . . ,  =                   ‹.           (7)
                              b 1 b 2     b n          a 1 a 2    a n          a 1 a 2     a n
                                                                          H n     ,  , . . . ,
                                                                               b 1 b 2     b n
            With the changes (4)-(7) in relation (1), it follows that

                 H n (a 1 , a 2 , . . . , a n )  1                  1                       1
                                    ≤                   ·                   ‹ ≤                     ‹,
                 H n (b 1 , b 2 , . . . , b n )  H n (b 1 , b 2 , . . . , b n )  1  1  1  a 1 a 2  a n
                                                          H n    ,   , . . . ,     H n    ,   , . . . ,
                                                               a 1 a 2     a n          b 1 b 2     b n
            that is the double inequality (3) from the statement (that implies the inequality (2)).

                But what happens if the sequences of numbers (a 1 , a 2 , . . . , a n ) and (b 1 , b 2 , . . . , b n ) are not
            reverse - ordered (as in the statements above)? For the case in which the sequences are similarly
            ordered, we have:

            Proposition 5. 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 similarly ordered, then
                                                    ‹                                              ‹‹
                                       1 1         1            A n (a 1 , a 2 , . . . , a n )  a 1 a 2  a n
               A n (a 1 , a 2 , . . . , a n ) · A n  ,  , . . . ,  ≥ max          , A n   ,   , . . . ,   .
                                       b 1 b 2    b n           A n (b 1 , b 2 , . . . , b n )  b 1 b 2  b n
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