Prove that ∑ i = 1 n ( x i x i + 1 ) 4 ≥ ∑ i = 1 n x i x i + 4 How to prove this inequality? ( x 1 x 2 ) 4 + ( x 2 x 3 ) 4 + ⋯ + ( x n x 1 ) 4 ≥ x 1 x 5 + x 2 x 6 + ⋯ + x n − 3 x 1 + x n − 2 x 2 + x n − 1 x 3 + x n x 4 ,where x 1 , x 2 , … , x n &gt; 0.

Question

Prove that       ∑          i      =      1        n              (                        x          i                          x                      i            +            1                              )              4        ≥      ∑          i      =      1        n              x      i              x              i        +        4            How to prove this inequality?            (                        x          1                          x          2                    )              4        +            (                        x          2                          x          3                    )              4        +  ⋯  +            (                        x          n                          x          1                    )              4        ≥            x      1              x      5        +            x      2              x      6        +  ⋯  +            x              n        −        3                    x      1        +            x              n        −        2                    x      2        +            x              n        −        1                    x      3        +            x      n              x      4        ,where       x    1    ,      x    2    ,  …  ,      x    n    &amp;gt;  0.

na1p1a2pafr · Accepted Answer

Let             x      1              x      2        =      y    1  ,             x      2              x      3        =      y    2  ,             x      3              x      4        =      y    3  ,             x      4              x      5        =      y    4  ,...Hence,       ∏          i      =      1        n        y    i    =            x      1              x              n        +        1             and we need to prove that      ∑          i      =      1        n        y    i    4    ≥      ∑          i      =      1        n        y    i        y          i      +      1            y          i      +      2            y          i      +      3        ,which is just AM-GM:      ∑          i      =      1        n        y    i    4    =      1    4        ∑          i      =      1        n    (      y    i    4    +      y          i      +      1        4    +      y          i      +      2        4    +      y          i      +      3        4    )  ≥      ∑          i      =      1        n        y    i        y          i      +      1            y          i      +      2            y          i      +      3      . Done!

How to prove this inequality? ((x_1)/(x_2))^(4)+((x_2)/(x_3))^{4}+...+((x_n)/(x_1))^(4)>=(x_1)/(x_5)+(x_2)/(x_6)+...+(x_(n-3))/(x_1)+(x_(n-2))/(x_2)+(x_(n-1))/(x_3)+(x_n)/(x_4), where x_1,x_2,…,x_n>0.

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