An Elementary Inequality ProblemGiven n positive real numbers x 1 , x 2 , . . . , x n + 1 , suppose: 1 1 + x 1 + 1 1 + x 2 + . . . + 1 1 + x n + 1 = 1Prove: x 1 x 2 . . . x n + 1 ≥ n n + 1 I have tried the substitution t i = 1 1 + x i The problem thus becomes:Given t 1 + t 2 + . . . + t n + 1 = 1 , t i &gt; 0Prove: ( 1 t 1 − 1 ) ( 1 t 2 − 1 ) . . . ( 1 t n + 1 − 1 ) ≥ n n + 1 Which is equavalent to the following: ( t 2 + t 3 + . . . + t n + 1 t 1 ) ( t 1 + t 3 + . . . + t n + 1 t 2 ) . . . ( t 1 + t 2 + . . . + t n t n + 1 ) ≥ n n + 1 From now on, I think basic equality (which says arithmetic mean is greater than geometric mean) can be applied but I cannot quite get there. Any hints? Thanks in advance.

Question

An Elementary Inequality ProblemGiven   n positive real numbers       x    1    ,      x    2    ,  .  .  .  ,      x          n      +      1      , suppose:      1          1      +              x        1              +      1          1      +              x        2              +  .  .  .  +      1          1      +              x                  n          +          1                      =  1Prove:      x    1        x    2    .  .  .      x          n      +      1        ≥            n              n        +        1            I have tried the substitution      t    i    =      1          1      +              x        i            The problem thus becomes:Given      t    1    +      t    2    +  .  .  .  +      t          n      +      1        =  1  ,      t    i    &amp;gt;  0Prove:  (      1          t      1        −  1  )  (      1          t      2        −  1  )  .  .  .  (      1          t              n        +        1              −  1  )  ≥            n              n        +        1            Which is equavalent to the following:  (                    t        2            +              t        3            +      .      .      .      +              t                  n          +          1                            t      1        )  (                    t        1            +              t        3            +      .      .      .      +              t                  n          +          1                            t      2        )  .  .  .  (                    t        1            +              t        2            +      .      .      .      +              t                  n                            t              n        +        1              )  ≥            n              n        +        1            From now on, I think basic equality (which says arithmetic mean is greater than geometric mean) can be applied but I cannot quite get there. Any hints? Thanks in advance.

frisiao · Accepted Answer

By AM-GM      ∏          i      =      1              n      +      1                  x              i                            x        i            +      1        =      ∏          i      =      1              n      +      1            ∑          k      ≠      i            1                  x        k            +      1        ≥      ∏          i      =      1              n      +      1            n                            ∏                      k            ≠            i                          (        1        +                  x          k                )            n        =            n              n        +        1                            ∏                  i          =          1                          n          +          1                    (      1      +              x        i            )      and we are done!

enmobladatn · Accepted Answer

From last, let                     t        1                    t        2            .      .      .              t                  n          +          1                      n    =  A with   A  M  −  G  M                    t        2            +              t        3            +      .      .      .      +              t                  n          +          1                      n    ≥                    t        2                    t        3            .      .      .              t                  n          +          1                      n    =      A                  t        1            n      so for the first      t    2    +      t    3    +  .  .  .  +      t          n      +      1        ≥            n      A                      t        1            n      and for the rest take like this.

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