Inequality x 1 2 x 1 2 + x 2 x 3 + x 2 2 x 2 2 + x 3 x 4 + ⋯ + x n − 1 2 x n − 1 2 + x n x 1 + x n 2 x n 2 + x 1 x 2 ≤ n − 1Show that for all n ≥ 2 x 1 2 x 1 2 + x 2 x 3 + x 2 2 x 2 2 + x 3 x 4 + ⋯ + x n − 1 2 x n − 1 2 + x n x 1 + x n 2 x n 2 + x 1 x 2 ≤ n − 1where x i are real positive numbersI was going to use x 1 2 x 1 2 + x 2 x 3 = 1 1 + x 2 x 3 x 1 2 ≤ x 1 2 ⋅ 1 x 2 x 3 ≤ x 1 4 ( 1 x 2 + 1 x 3 ) x 1 2 x 1 2 + x 2 x 3 + x 2 2 x 2 2 + x 3 x 4 + . . . + x n − 2 2 x n − 2 2 + x n − 1 x n + x n − 1 2 x n − 1 2 + x n x 1 + x n 2 x n 2 + x 1 x 2 = 1 1 + x 2 x 3 x 1 2 + 1 1 + x 3 x 4 x 2 2 + . . . + 1 1 + x n − 1 x n x n − 2 2 + 1 1 + x n x 1 x n − 1 2 + 1 1 + x 1 x 2 x n 2 ≤ x 1 4 ( 1 x 2 + 1 x 3 ) + x 2 4 ( 1 x 3 + 1 x 4 ) + . . . + x n − 2 4 ( 1 x n − 1 + 1 x n ) + x n − 1 4 ( 1 x n + 1 x 1 ) + x n 4 ( 1 x 1 + 1 x 2 ) = 1 4 ( ( x 1 x 2 + x 1 x 3 ) + ( x 2 x 3 + x 2 x 4 ) + ( x 3 x 4 + x 3 x 5 ) + . . . + ( x n − 2 x n − 1 + x n − 2 x n ) + ( x n − 1 x n + x n − 1 x 1 ) + ( x n x 1 + x n x 2 ) ) = 1 4 ( ( x 1 + x 2 x 3 ) + ( x 2 + x 3 x 4 ) + ( x 3 + x 4 x 5 ) + . . . + ( x n − 3 + x n − 2 x n − 1 ) + ( x n − 1 + x n − 2 x n ) + ( x 1 + x n x 2 ) + ( x n − 1 + x n x 1 ) ) ....I thought about using Cauchy's inequality, but that would only increase the problem

Question

Inequality             x      1      2                      x        1        2            +              x        2                    x        3              +            x      2      2                      x        2        2            +              x        3                    x        4              +  ⋯  +            x              n        −        1            2                      x                  n          −          1                2            +              x        n                    x        1              +            x      n      2                      x        n        2            +              x        1                    x        2              ≤  n  −  1Show that for all   n  ≥  2            x      1      2                      x        1        2            +              x        2                    x        3              +            x      2      2                      x        2        2            +              x        3                    x        4              +  ⋯  +            x              n        −        1            2                      x                  n          −          1                2            +              x        n                    x        1              +            x      n      2                      x        n        2            +              x        1                    x        2              ≤  n  −  1where       x    i   are real positive numbersI was going to use             x              1                    2                            x                  1                          2                    +              x                  2                            x                  3                      =      1          1      +                                    x                          2                                            x                          3                                                x                      1                                2                                ≤            x              1              2    ⋅      1                  x                  2                            x                  3                      ≤            x              1              4        (          1              x                  2                      +          1              x                  3                      )              x              1                    2                            x                  1                          2                    +              x                  2                            x                  3                      +            x              2                    2                            x                  2                          2                    +              x                  3                            x                  4                      +  .  .  .  +            x              n        −        2                    2                            x                  n          −          2                          2                    +              x                  n          −          1                            x                  n                      +            x              n        −        1                    2                            x                  n          −          1                          2                    +              x                  n                            x                  1                      +            x              n                    2                            x                  n                          2                    +              x                  1                            x                  2                      =      1          1      +                                    x                          2                                            x                          3                                                x                      1                                2                                +      1          1      +                                    x                          3                                            x                          4                                                x                      2                                2                                +  .  .  .  +      1          1      +                                    x                          n              −              1                                            x                          n                                                x                      n            −            2                                2                                +      1          1      +                                    x                          n                                            x                          1                                                x                      n            −            1                                2                                +      1          1      +                                    x                          1                                            x                          2                                                x                      n                                2                                ≤            x              1              4        (          1              x                  2                      +          1              x                  3                      )    +            x              2              4        (          1              x                  3                      +          1              x                  4                      )    +  .  .  .  +            x              n        −        2              4        (          1              x                  n          −          1                      +          1              x                  n                      )    +            x              n        −        1              4        (          1              x                  n                      +          1              x                  1                      )    +            x              n              4        (          1              x                  1                      +          1              x                  2                      )    =      1    4        (          (                        x                      1                                    x                      2                              +                        x                      1                                    x                      3                              )        +          (                        x                      2                                    x                      3                              +                        x                      2                                    x                      4                              )        +          (                        x                      3                                    x                      4                              +                        x                      3                                    x                      5                              )        +    .    .    .    +          (                        x                      n            −            2                                    x                      n            −            1                              +                        x                      n            −            2                                    x                      n                              )        +          (                        x                      n            −            1                                    x                      n                              +                        x                      n            −            1                                    x                      1                              )        +          (                        x                      n                                    x                      1                              +                        x                      n                                    x                      2                              )        )    =      1    4        (          (                                    x                          1                                +                      x                          2                                                x                      3                              )        +          (                                    x                          2                                +                      x                          3                                                x                      4                              )        +          (                                    x                          3                                +                      x                          4                                                x                      5                              )        +    .    .    .    +          (                                    x                          n              −              3                                +                      x                          n              −              2                                                x                      n            −            1                              )        +          (                                    x                          n              −              1                                +                      x                          n              −              2                                                x                      n                              )        +          (                                    x                          1                                +                      x                          n                                                x                      2                              )        +          (                                    x                          n              −              1                                +                      x                          n                                                x                      1                              )        )  ....I thought about using Cauchy&#039;s inequality, but that would only increase the problem

Kyle Delacruz · Accepted Answer

Let                     x        2                    x        3                    x      1      2        =            a      1              a      2      ,... and similar, where       a    i    &amp;gt;  0 and       a          n      +      1        =      a    1  Thus, we need to prove that:      ∑          i      =      1        n        1          1      +                        a          i                          a                      i            +            1                                ≤  n  −  1or      ∑          i      =      1        n        (          1              1        +                              a            i                                a                          i              +              1                                            −    1    )    ≤  −  1or      ∑          i      =      1        n              a      i                      a        i            +              a                  i          +          1                      ≥  1  ,which is true because      ∑          i      =      1        n              a      i                      a        1            +              a                  i          +          1                      ≥      ∑          i      =      1        n              a      i                      a        1            +              a        2            +      .      .      .      +              a        n              =  1

Show that for all n>=2 (x_1^2)/(x_1^2+x_2x_3)+(x_2^2)/(x_2^2+x_3x_4)+...+(x_(n-1}^2)/(x_(n-1)^2+x_nx_1)+(x_n^2)/(x_n^2+x_1x_2)<= n-1 where x_i are real positive numbers

Answered question

Answer & Explanation

New Questions in Pre-Algebra