Is anyway to prove this: ∏ k = 1 n ( a k ) &lt; ( 1 / n n ) ∗ ( ∑ k = 1 n ( 1 + a k ∗ a k + 1 ) ) n ak and n are positive real number greater than 0.EDIT: a_{k+1} becomes a_{1} when a_{k}=a_{n}, it is a cylic notation. SORRY.Any ideas of how to attack the problem?? Thank You.I don't know if this could help, but the 1/n is also the exponent for the left hand side. I'm thinking maybe of log??I'm pretty sure that at some point it would be helpful the binominal coefficent?? I don't know.

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Is anyway to prove this:       ∏          k      =      1              n        (      a          k        )  &amp;lt;  (  1      /        n    n    )  ∗  (      ∑          k      =      1              n        (      1    +          a              k              ∗          a              k        +        1              )      )    n  ak and n are positive real number greater than 0.EDIT: a_{k+1} becomes a_{1} when a_{k}=a_{n}, it is a cylic notation. SORRY.Any ideas of how to attack the problem?? Thank You.I don&#039;t know if this could help, but the 1/n is also the exponent for the left hand side. I&#039;m thinking maybe of log??I&#039;m pretty sure that at some point it would be helpful the binominal coefficent?? I don&#039;t know.

Christina Ward · Accepted Answer

After the question get cleared up, the answer becomes a trivial application of GM   ≤ AM. Since the       a    k   are in cyclic notation. i.e.       a          n      +      1        =      a    1  , we have                              ∏                      k            =            1                    n                          a          k                                    =                  ∏                      k            =            1                    n                                      a            k                                a            k                          =                              a            1                                    (                      ∏                          k              =              1                                      n              −              1                                                          a              k                                      a                              k                +                1                                              )                                      a            n                                                            =                  ∏                      k            =            1                    n                                      a            k                                a                          k              +              1                                                        ⟵                      a                          1                                =                      a                          n              +              1                                 and rearrange                                                   ≤                              (                                                            ∑                                      k                    =                    1                                    n                                                                      a                    k                                                        a                                          k                      +                      1                                                                                  n                        )                    n                                  ⟵          GM          ≤          AM                                                  &amp;lt;                              (                                                            ∑                                      k                    =                    1                                    n                                                  1                  +                                      a                    k                                                        a                                          k                      +                      1                                                                                  n                        )                    n                =                  1                      n            n                                                (                          ∑                              k                =                1                            n                                      1              +                              a                k                                            a                                  k                  +                  1                                                      )                    n

Is anyway to prove this: <munderover> ∏ <mrow class="MJX-TeXAtom-ORD"

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