Prove the inequality a 2 + b 2 c 2 + a + b + b 2 + c 2 a 2 + b + c + c 2 + a 2 b 2 + c + a ≥ 2For every a , b , c &gt; 0 such that a b c = 1 prove the inequalityMy work so far: a b c = 1 ⇒ 1 3 ( a + b + c ) ≥ 1 Then c 2 + a + b ≤ c 2 + ( a + b ) 1 3 ( a + b + c )Or more ( a 2 + b 2 c 2 + a + b + b 2 + c 2 a 2 + b + c + c 2 + a 2 b 2 + c + a ) ( ( c 2 + a + b ) + ( a 2 + b + c ) + ( b 2 + c + a ) ) ≥ ≥ ( a 2 + b 2 + b 2 + c 2 + c 2 + a 2 ) 2

Question

Prove the inequality                     a        2            +              b        2                            c        2            +      a      +      b        +                    b        2            +              c        2                            a        2            +      b      +      c        +                    c        2            +              a        2                            b        2            +      c      +      a        ≥  2For every   a  ,  b  ,  c  &amp;gt;  0 such that   a  b  c  =  1 prove the inequalityMy work so far:  a  b  c  =  1  ⇒      1    3    (  a  +  b  +  c  )  ≥  1 Then      c    2    +  a  +  b  ≤      c    2    +  (  a  +  b  )      1    3    (  a  +  b  +  c  )Or more  (                    a        2            +              b        2                            c        2            +      a      +      b        +                    b        2            +              c        2                            a        2            +      b      +      c        +                    c        2            +              a        2                            b        2            +      c      +      a        )  (  (      c    2    +  a  +  b  )  +  (      a    2    +  b  +  c  )  +  (      b    2    +  c  +  a  )  )  ≥  ≥  (            a      2        +          b      2        +            b      2        +          c      2        +            c      2        +          a      2            )    2

Bria Berg · Accepted Answer

Let   a  =      x    3  ,   b  =      y    3   and   c  =      z    3  . Hence, by Muirhead       ∑          c      y      c                          a        2            +              b        2                            c        2            +      a      +      b        =      ∑          c      y      c                          x        6            +              y        6                            z        6            +      x      y      z      (              x        3            +              y        3            )        ≥      ∑          c      y      c                          x        6            +              y        6                            z        6            +      z      (              x        5            +              y        5            )        =                    ∑                  c          y          c                    (              x        7            y      +              x        7            z      )              x      y      z      (              x        5            +              y        5            +              z        5            )        ≥  2

3c4ar1bzki1u · Accepted Answer

As we know:  2  a  b  ≤      a          2        +      b          2      , and  3      a                  1        3              =  3  (      a          2        b  c      )                  1        3              ≤      a          2        +  b  +  c(because of   a  b  c  =  1)Now we have :  2  (            a      b              3              c                  1                      /                    3                      +            a      c              3              b                  1                      /                    3                      +            c      b              3              a                  1                      /                    3                      )Now using Cauchy we could prove our inequality.

Prove the inequality a 2 </msup> + b

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