For a , b , c &gt; and a 1 + a + 2 b 1 + b + 3 c 1 + c ≤ 1. Prove that a b 2 c 3 d 5 ≤ 1 10 11 a 1 + a + 2 b 1 + b + 3 c 1 + c ≤ 1 ↔ 1 1 + a + 2 1 + b + 3 1 + c ≥ 5 =&gt; 1 a + 1 ≥ 2 − 2 1 + b + 3 − 3 1 + b = 2 b 1 + b + 3 c 1 + c ≥ 5 b 2 ( 1 + b ) 2 ∗ c 3 ( 1 + c ) 3 5 1 1 + b ≥ a 1 + a + b 1 + b + 3 c 1 + c ≥ 5 a 1 + a ∗ b 1 + b ∗ c 3 ( 1 + c ) 3 5 1 1 + c ≥ a 1 + a + 2 b 1 + b + 2 c 1 + c ≥ 5 a 1 + a . b 2 ( 1 + b ) 2 . c 2 ( 1 + c ) 2 5 =&gt; 1 1 + a ∗ ( 1 1 + b ) 2 ∗ ( 1 1 + c ) 3 ≥ 5 6 a ( b 2 ) ( c 3 ) ( 1 + a ) ( 1 + b ) 2 ( 1 + c ) 3 Seem wrong, fix for me

Question

For   a  ,  b  ,  c  &amp;gt; and       a          1      +      a        +            2      b              1      +      b        +            3      c              1      +      c        ≤  1. Prove that  a      b    2        c    3        d    5    ≤      1          10              11                  a          1      +      a        +            2      b              1      +      b        +            3      c              1      +      c        ≤  1  ↔      1          1      +      a        +      2          1      +      b        +      3          1      +      c        ≥  5  =&amp;gt;      1          a      +      1        ≥  2  −      2          1      +      b        +  3  −      3          1      +      b        =            2      b              1      +      b        +            3      c              1      +      c        ≥  5                              b          2                          (          1          +          b                      )            2                              ∗                        c          3                          (          1          +          c                      )            3                                5        1          1      +      b        ≥      a          1      +      a        +      b          1      +      b        +            3      c              1      +      c        ≥  5                    a                  1          +          a                    ∗              b                  1          +          b                    ∗                        c          3                          (          1          +          c                      )            3                                5        1          1      +      c        ≥      a          1      +      a        +            2      b              1      +      b        +            2      c              1      +      c        ≥  5                    a                  1          +          a                    .                        b          2                          (          1          +          b                      )            2                              .                        c          2                          (          1          +          c                      )            2                                5    =&amp;gt;      1          1      +      a        ∗      (          1              1        +        b                    )      2        ∗  (      1          1      +      c            )    3    ≥      5    6              a      (              b        2            )      (              c        3            )              (              1        +        a            )            (          1      +      b              )      2                          (                  1          +          c                )            3      Seem wrong, fix for me

suponeriq · Accepted Answer

I think it means the following inequality.Let   a,   b,   c and   d be positive numbers such that      a          a      +      1        +            2      b              b      +      1        +            3      c              c      +      1        +            5      d              d      +      1        ≤  1.Prove that  a      b    2        c    3        d    5    ≤      1          10              11              .I think your idea works here very well.From the condition by AM-GM we obtain:      1          1      +      a        ≥            2      b              b      +      1        +            3      c              c      +      1        +            5      d              d      +      1        ≥  10                              b          2                          c          3                          d          5                            (        1        +        b                  )          2                (        1        +        c                  )          3                (        1        +        d                  )          5                      10    ;      1          1      +      b        ≥      a          1      +      a        +      b          b      +      1        +            3      c              c      +      1        +            5      d              d      +      1        ≥  10                    a        b                  c          3                          d          5                            (        1        +        a        )        (        1        +        b        )        (        1        +        c                  )          3                (        1        +        d                  )          5                      10    ;      1          1      +      c        ≥      a          1      +      a        +            2      b              b      +      1        +            2      c              c      +      1        +            5      d              d      +      1        ≥  10                    a                  b          2                          c          2                          d          5                            (        1        +        a        )        (        1        +        b                  )          2                (        1        +        c                  )          2                (        1        +        d                  )          5                      10  and      1          1      +      d        ≥      a          a      +      1        +            2      b              b      +      1        +            3      c              c      +      1        +            4      d              d      +      1        ≥  10                    a                  b          2                          c          3                          d          4                            (        1        +        a        )        (        1        +        b                  )          2                (        1        +        c                  )          3                (        1        +        d                  )          4                      10    .Thus,      1          (      1      +      a      )      (      1      +      b              )        2            (      1      +      c              )        3            (      1      +      d              )        5              ≥      10          1      +      2      +      3      +      5                                    a                      2            +            3            +            5                                    b                      2            +            2            +            6            +            10                                    c                      3            +            6            +            6            +            15                                    d                      5            +            10            +            15            +            20                                      (        1        +        a                  )                      10                          (        1        +        b                  )                      20                          (        1        +        c                  )                      30                          (        1        +        d                  )                      50                                10  or  a      b    2        c    3        d    5    ≤      1          10              11            and we are done!

For a,b,c> and (a)/(1+a)+(2b)/(1+b)+(3c)/(1+c)<=1 . Prove that ab^2c^3d^5<=(1)/(10^(11))

Answered question

Answer & Explanation

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