Prove that a b + 2 c + 3 d + b c + 2 d + 3 a + c d + 2 a + 3 b + d a + 2 b + 3 c ≥ 2 3 .I was thinking of trying a ≥ b ≥ c since the inequality is cyclic. I am not sure how to use this, though, or if this would help simplify the inequality. It also doesn't look like I can really use AM-GM or Cauchy-Schwarz.

Question

Prove that       a          b      +      2      c      +      3      d        +      b          c      +      2      d      +      3      a        +      c          d      +      2      a      +      3      b        +      d          a      +      2      b      +      3      c        ≥      2    3    .I was thinking of trying   a  ≥  b  ≥  c since the inequality is cyclic. I am not sure how to use this, though, or if this would help simplify the inequality. It also doesn&#039;t look like I can really use AM-GM or Cauchy-Schwarz.

lofoptiformfp · Accepted Answer

We need only note that                                             a                      b            +            2            c            +            3            d                          +                  b                      c            +            2            d            +            3            a                          +                  c                      d            +            2            a            +            3            b                          +                  d                      a            +            2            b            +            3            c                                              =                                                   a            2                                a            b            +            2            a            c            +            3            a            d                          +                              b            2                                b            c            +            2            b            d            +            3            a            b                          +                              c            2                                c            d            +            2            a            c            +            3            b            c                          +                              d            2                                a            d            +            2            b            d            +            3            c            d                                              ≥                                                   (            a            +            b            +            c            +            d                          )              2                                            4            (            a            b            +            a            c            +            a            d            +            b            c            +            b            d            +            c            d            )                                              =                                                                 a              2                        +                          b              2                        +                          c              2                        +                          d              2                        +            2            (            a            b            +            a            c            +            a            d            +            b            c            +            b            d            +            c            d            )                                4            (            a            b            +            a            c            +            a            d            +            b            c            +            b            d            +            c            d            )                                              ≥                                                                 2              3                        (            a            b            +            a            c            +            a            d            +            b            c            +            b            d            +            c            d            )            +            2            (            a            b            +            a            c            +            a            d            +            b            c            +            b            d            +            c            d            )                                4            (            a            b            +            a            c            +            a            d            +            b            c            +            b            d            +            c            d            )                                              =                                       2          3                .            EDIT:                    (        a        +        b        +        c        +        d                  )          2                                    =                              (                                    a                      a            b            +            2            a            c            +            3            a            d                                    a          b          +          2          a          c          +          3          a          d                                                          +                  b                      b            c            +            2            b            d            +            3            a            b                                    b          c          +          2          b          d          +          3          a          b                                                          +                  c                      c            d            +            2            a            c            +            3            b            c                                    c          d          +          2          a          c          +          3          b          c                                                          +                  d                      a            d            +            2            b            d            +            3            c            d                                    a          d          +          2          b          d          +          3          c          d                                                    )                                2                                                  ≤                              (                                                a            2                                a            b            +            2            a            c            +            3            a            d                          +                              b            2                                b            c            +            2            b            d            +            3            a            b                          +                              c            2                                c            d            +            2            a            c            +            3            b            c                                                                    +                              d            2                                a            d            +            2            b            d            +            3            c            d                                                )                                                [                          4        (        a        b        +        a        c        +        a        d        +        b        c        +        b        d        +        c        d        )                              ]                          .            That is,                                                         a            2                                a            b            +            2            a            c            +            3            a            d                          +                              b            2                                b            c            +            2            b            d            +            3            a            b                          +                              c            2                                c            d            +            2            a            c            +            3            b            c                          +                              d            2                                a            d            +            2            b            d            +            3            c            d                                              ≥                                                   (            a            +            b            +            c            +            d                          )              2                                            4            (            a            b            +            a            c            +            a            d            +            b            c            +            b            d            +            c            d            )                          .            Moreover,                    (        a        +        b        +        c        +        d                  )          2                                    =                  a          2                +                  b          2                +                  c          2                +                  d          2                +        2        (        a        b        +        a        c        +        a        d        +        b        c        +        b        d        +        c        d        )                                          =                  1          3                                      [                                                (                                    a          2                +                  b          2                                      )                          +                              (                                    a          2                +                  c          2                                      )                          +                              (                                    a          2                +                  d          2                                      )                          +                              (                                    b          2                +                  c          2                                      )                          +                              (                                    b          2                +                  d          2                )                              )                                                                    +                              (                                    c          2                +                  d          2                                      )                                                ]                          +        2        (        a        b        +        a        c        +        a        d        +        b        c        +        b        d        +        c        d        )                                          ≥                  2          3                (        a        b        +        a        c        +        a        d        +        b        c        +        b        d        +        c        d        )        +        2        (        a        b        +        a        c        +        a        d        +        b        c        +        b        d        +        c        d        )                                          =                  8          3                (        a        b        +        a        c        +        a        d        +        b        c        +        b        d        +        c        d        )        .

spockmonkey40 · Accepted Answer

By C-S       ∑          c      y      c            a          b      +      2      c      +      3      d        ≥            (      a      +      b      +      c      +      d              )        2                            ∑                  c          y          c                    (      4      a      b      +      2      a      c      )        ≥      2    3  , where the last inequality it&#039;s      ∑          s      y      m        (  a  −  b      )    2    ≥  0. Done!

Prove that a b + 2 c + 3 d </m

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