Inequality Challenge with a + b + c = 3I tried to prove this inequality, but I failed. It seemed that if one uses Cauchy-Schwarz inequality on the right hand side, it may expand RHS's value larger than LHS's. Problem is presented as follows:If a , b , c ≥ 0 and a + b + c = 3, prove that 2 ( a b + b c + c a ) − 3 a b c ≥ a ⋅ b 2 + c 2 2 + b ⋅ c 2 + a 2 2 + c ⋅ a 2 + b 2 2

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Inequality Challenge with   a  +  b  +  c  =  3I tried to prove this inequality, but I failed. It seemed that if one uses Cauchy-Schwarz inequality on the right hand side, it may expand RHS&#039;s value larger than LHS&#039;s. Problem is presented as follows:If   a  ,  b  ,  c  ≥  0 and   a  +  b  +  c  =  3, prove that  2  (  a  b  +  b  c  +  c  a  )  −  3  a  b  c  ≥  a  ⋅                              b          2                +                  c          2                    2        +  b  ⋅                              c          2                +                  a          2                    2        +  c  ⋅                              a          2                +                  b          2                    2

cegielnikmzjkf · Accepted Answer

After using                               a          2                +                  b          2                    2        ≤            3              a        2            +      2      a      b      +      3              b        2                    4      (      a      +      b      )      which is just   (  a  −  b      )    4    ≥  0, you&#039;ll get something obvious.Indeed, it remains to prove that  2  (  a  b  +  a  c  +  b  c  )  −            9      a      b      c              a      +      b      +      c        ≥      ∑          c      y      c                  c      (      3              a        2            +      2      a      b      +      3              b        2            )              4      (      a      +      b      )      or  2  (  a  b  +  a  c  +  b  c  )  −            9      a      b      c              a      +      b      +      c        ≥      3    2    (  a  b  +  a  c  +  b  c  )  −  a  b  c      ∑          c      y      c            1          a      +      b      or      1    2    (  a  b  +  a  c  +  b  c  )  −            9      a      b      c              2      (      a      +      b      +      c      )        −            9      a      b      c              2      (      a      +      b      +      c      )        +  a  b  c      ∑          c      y      c            1          a      +      b        ≥  0or      ∑          c      y      c        c  (  a  −  b      )    2    +  a  b  c      (          ∑              c        y        c              (    a    +    b    )          ∑              c        y        c                    1              a        +        b              −    9    )    ≥  0Done!

Inequality Challenge with a + b + c = 3 I tried to prove this inequality, b

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