If a , b , c ∈ ( 0 , ∞ ) with abc=1 prove: a 2 + b 2 + c 2 + 3 ( a b + b c + c a ) ≤ 1 2 ( a + b + c ) ( a + b ) ( b + c ) ( c + a ) .On atypical inequality (a smaller amount than a product) that resisted my attempts to prove it.

Question

If   a  ,  b  ,  c  ∈  (  0  ,  ∞  ) with abc=1 prove:      a    2    +      b    2    +      c    2    +  3  (  a  b  +  b  c  +  c  a  )  ≤      1    2    (      a    +      b    +      c    )  (  a  +  b  )  (  b  +  c  )  (  c  +  a  )  .On atypical inequality (a smaller amount than a product) that resisted my attempts to prove it.

Braeden Shannon · Accepted Answer

We need to prove that      ∏          c      y      c        (      a    2    +      b    2    )      ∑          c      y      c        a  ≥  2  a  b  c      ∑          c      y      c        (      a    4    +  3      a    2        b    2    )for positives a, b and c or      ∑          c      y      c            (          a      4              b      2        +          a      4              c      2        +          2      3              a      2              b      2              c      2        )        ∑          c      y      c        a  ≥  2      ∑          c      y      c        (      a    5    b  c  +  3      a    3        b    3    c  )or      ∑          c      y      c        (      a    5        b    2    +      a    5        c    2    +      a    4        b    3    +      a    4        c    3    +      a    4        b    2    c  +      a    4        c    2    b  +  2      a    3        b    2        c    2    −  2      a    5    b  c  −  6      a    3        b    3    c  )  ≥  0  ,which is true because      ∑          c      y      c        (      a    5        b    2    +      a    5        c    2    −  2      a    5    b  c  )  =      ∑          c      y      c            a    5    (  b  −  c      )    2    ≥  0and by AM-GM      ∑          c      y      c        (      a    4        b    3    +      a    4        c    3    +      a    4        b    2    c  +      a    4        c    2    b  +  2      a    3        b    2        c    2    −  6      a    3        b    3    c  )  =  =      ∑          c      y      c        (      a    4        b    3    +      a    3        b    4    +      a    4        b    2    c  +      b    4        a    2    c  +      a    3        b    2        c    2    +      b    3        a    2        c    2    −  6      a    3        b    3    c  )  ≥  ≥      ∑          c      y      c            (    6                            a          4                          b          3                ⋅                  a          3                          b          4                ⋅                  a          4                          b          2                c        ⋅                  b          4                          a          2                c        ⋅                  a          3                          b          2                          c          2                ⋅                  b          3                          a          2                          c          2                    6        −    6          a      3              b      3        c    )    =  0.Done!

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