For a , b , c. Prove that a 2 a + b c 3 + b 2 b + c a 3 + c 2 c + a b 3 ≥ 3 2 By Cauchy-Schwarz: ⇒ a 2 a + b c 3 + b 2 b + c a 3 + c 2 c + a b 3 ≥ ( a + b + c ) 2 a + b + c + b c 3 + c a 3 + a b 3 Hence we need prove 9 3 + b c 3 + c a 3 + a b 3 ≥ 3 2 ( ∗ ) ⇔ 9 ≥ 3 a b 3 + 3 b c 3 + 3 c a 3 By AM-GM a + b + 1 ≥ 3 a b 3 and . . . ⇒ 2 ( a + b + c ) + 3 ≥ 3 a b 3 + 3 b c 3 + 3 c a 3 ⇒ 9 ≥ 3 a b 3 + 3 b c 3 + 3 c a 3 I need a new method.

Question

For   a  ,  b  ,  c. Prove that             a      2              a      +                        b          c                3              +            b      2              b      +                        c          a                3              +            c      2              c      +                        a          b                3              ≥      3    2  By Cauchy-Schwarz:   ⇒                    a        2                    a        +                              b            c                    3                      +                    b        2                    b        +                              c            a                    3                      +                    c        2                    c        +                              a            b                    3                      ≥                              (          a          +          b          +          c          )                2                    a        +        b        +        c        +                              b            c                    3                +                              c            a                    3                +                              a            b                    3                    Hence we need prove             9              3        +                              b            c                    3                +                              c            a                    3                +                              a            b                    3                      ≥            3      2        (  ∗  )  ⇔  9  ≥  3            a      b        3    +  3            b      c        3    +  3            c      a        3  By AM-GM   a  +  b  +  1  ≥  3            a      b        3   and   .  .  .  ⇒  2      (    a    +    b    +    c    )    +  3  ≥  3            a      b        3    +  3            b      c        3    +  3            c      a        3    ⇒  9  ≥  3            a      b        3    +  3            b      c        3    +  3            c      a        3  I need a new method.

Belen Solomon · Accepted Answer

By AM-GM and C-S we obtain:      ∑          c      y      c                  a      2              a      +                        b          c                3              ≥      ∑          c      y      c                  a      2              a      +                        b          +          c          +          1                3              =      ∑          c      y      c                  9              a        2                    3      (      3      a      +      b      +      c      )      +      a      +      b      +      c        =  =      ∑          c      y      c                  9              a        2                    10      a      +      4      b      +      4      c        ≥            9      (      a      +      b      +      c              )        2                            ∑                  c          y          c                    (      10      a      +      4      b      +      4      c      )        =            9      (      a      +      b      +      c              )        2                    18      (      a      +      b      +      c      )        =      3    2

For a,b,c are positive real number satisfy a+b+c=3. Prove that (a^2)/(a+root[3]{bc))+(b^2)/(b+root[3](ca))+(c^2)/(c+root[3](ab))>=(3)/(2)

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