Clarification about a proofit's just a clarification about a proof :The starting inequality is : 0.5 ≥ 1.5 − 1 a 2 + a b + 2 a + b 2 + 3 a 2 + a b − 2 a + b 2 + 3 + 1 − 1 c 2 + c b + 2 b + b 2 + 3 c 2 + c b − 2 b + b 2 + 3 + 1 − 1 a 2 + a c + 2 c + c 2 + 3 a 2 + a c − 2 c + c 2 + 3 + 1 So we study the function f ( x ) = 0.5 − 1 x + 1 wich is concave so we can apply Jensen inequality we get : 0.5 − 1 a 2 + a b + 2 a + b 2 + 3 a 2 + a b − 2 a + b 2 + 3 + c 2 + c b + 2 b + b 2 + 3 c 2 + c b − 2 b + b 2 + 3 + a 2 + a c + 2 c + c 2 + 3 a 2 + a c − 2 c + c 2 + 3 3 + 1 ≥ 1 3 ( 1.5 − 1 a 2 + a b + 2 a + b 2 + 3 a 2 + a b − 2 a + b 2 + 3 + 1 − 1 c 2 + c b + 2 b + b 2 + 3 c 2 + c b − 2 b + b 2 + 3 + 1 − 1 a 2 + a c + 2 c + c 2 + 3 a 2 + a c − 2 c + c 2 + 3 + 1 )Wich is equivalent to : 1.5 − 3 a 2 + a b + 2 a + b 2 + 3 a 2 + a b − 2 a + b 2 + 3 + c 2 + c b + 2 b + b 2 + 3 c 2 + c b − 2 b + b 2 + 3 + a 2 + a c + 2 c + c 2 + 3 a 2 + a c − 2 c + c 2 + 3 3 + 1 ≥ ( 1.5 − 1 a 2 + a b + 2 a + b 2 + 3 a 2 + a b − 2 a + b 2 + 3 + 1 − 1 c 2 + c b + 2 b + b 2 + 3 c 2 + c b − 2 b + b 2 + 3 + 1 − 1 a 2 + a c + 2 c + c 2 + 3 a 2 + a c − 2 c + c 2 + 3 + 1 )But it's easy to find the minimum of : 3 a 2 + a b + 2 a + b 2 + 3 a 2 + a b − 2 a + b 2 + 3 + c 2 + c b + 2 b + b 2 + 3 c 2 + c b − 2 b + b 2 + 3 + a 2 + a c + 2 c + c 2 + 3 a 2 + a c − 2 c + c 2 + 3 3 + 1 Wich is one so we have : 0.5 ≥ 1.5 − 1 a 2 + a b + 2 a + b 2 + 3 a 2 + a b − 2 a + b 2 + 3 + 1 − 1 c 2 + c b + 2 b + b 2 + 3 c 2 + c b − 2 b + b 2 + 3 + 1 − 1 a 2 + a c + 2 c + c 2 + 3 a 2 + a c − 2 c + c 2 + 3 + 1 Done !

Question

Clarification about a proofit&#039;s just a clarification about a proof :The starting inequality is :  0.5  ≥  1.5  −      1                                        a            2                    +          a          b          +          2          a          +                      b            2                    +          3                                      a            2                    +          a          b          −          2          a          +                      b            2                    +          3                    +      1        −      1                                        c            2                    +          c          b          +          2          b          +                      b            2                    +          3                                      c            2                    +          c          b          −          2          b          +                      b            2                    +          3                    +      1        −      1                                        a            2                    +          a          c          +          2          c          +                      c            2                    +          3                                      a            2                    +          a          c          −          2          c          +                      c            2                    +          3                    +      1      So we study the function   f  (  x  )  =  0.5  −      1          x      +      1       wich is concave so we can apply Jensen inequality we get :  0.5  −      1                                                                      a                2                            +              a              b              +              2              a              +                              b                2                            +              3                                                      a                2                            +              a              b              −              2              a              +                              b                2                            +              3                                +                                                    c                2                            +              c              b              +              2              b              +                              b                2                            +              3                                                      c                2                            +              c              b              −              2              b              +                              b                2                            +              3                                +                                                                      a                  2                                +                a                c                +                2                c                +                                  c                  2                                +                3                                                              a                  2                                +                a                c                −                2                c                +                                  c                  2                                +                3                                                    3            +      1        ≥      1    3    (  1.5  −      1                                        a            2                    +          a          b          +          2          a          +                      b            2                    +          3                                      a            2                    +          a          b          −          2          a          +                      b            2                    +          3                    +      1        −      1                                        c            2                    +          c          b          +          2          b          +                      b            2                    +          3                                      c            2                    +          c          b          −          2          b          +                      b            2                    +          3                    +      1        −      1                                        a            2                    +          a          c          +          2          c          +                      c            2                    +          3                                      a            2                    +          a          c          −          2          c          +                      c            2                    +          3                    +      1        )Wich is equivalent to :  1.5  −      3                                                                      a                2                            +              a              b              +              2              a              +                              b                2                            +              3                                                      a                2                            +              a              b              −              2              a              +                              b                2                            +              3                                +                                                    c                2                            +              c              b              +              2              b              +                              b                2                            +              3                                                      c                2                            +              c              b              −              2              b              +                              b                2                            +              3                                +                                                                      a                  2                                +                a                c                +                2                c                +                                  c                  2                                +                3                                                              a                  2                                +                a                c                −                2                c                +                                  c                  2                                +                3                                                    3            +      1        ≥  (  1.5  −      1                                        a            2                    +          a          b          +          2          a          +                      b            2                    +          3                                      a            2                    +          a          b          −          2          a          +                      b            2                    +          3                    +      1        −      1                                        c            2                    +          c          b          +          2          b          +                      b            2                    +          3                                      c            2                    +          c          b          −          2          b          +                      b            2                    +          3                    +      1        −      1                                        a            2                    +          a          c          +          2          c          +                      c            2                    +          3                                      a            2                    +          a          c          −          2          c          +                      c            2                    +          3                    +      1        )But it&#039;s easy to find the minimum of :      3                                                                      a                2                            +              a              b              +              2              a              +                              b                2                            +              3                                                      a                2                            +              a              b              −              2              a              +                              b                2                            +              3                                +                                                    c                2                            +              c              b              +              2              b              +                              b                2                            +              3                                                      c                2                            +              c              b              −              2              b              +                              b                2                            +              3                                +                                                                      a                  2                                +                a                c                +                2                c                +                                  c                  2                                +                3                                                              a                  2                                +                a                c                −                2                c                +                                  c                  2                                +                3                                                    3            +      1      Wich is one so we have :  0.5  ≥  1.5  −      1                                        a            2                    +          a          b          +          2          a          +                      b            2                    +          3                                      a            2                    +          a          b          −          2          a          +                      b            2                    +          3                    +      1        −      1                                        c            2                    +          c          b          +          2          b          +                      b            2                    +          3                                      c            2                    +          c          b          −          2          b          +                      b            2                    +          3                    +      1        −      1                                        a            2                    +          a          c          +          2          c          +                      c            2                    +          3                                      a            2                    +          a          c          −          2          c          +                      c            2                    +          3                    +      1      Done !

Yasmin Lam · Accepted Answer

You proved that      ∑          c      y      c            1                                        a            2                    +          a          b          +          2          a          +                      b            2                    +          3                                      a            2                    +          a          b          −          2          a          +                      b            2                    +          3                    +      1        ≥      3                                        ∑                          c              y              c                                                                          a                2                            +              a              b              +              2              a              +                              b                2                            +              3                                                      a                2                            +              a              b              −              2              a              +                              b                2                            +              3                                      3            +      1      or      ∑          c      y      c                          a        2            +      a      b      −      2      a      +              b        2            +      3                      a        2            +      a      b      +              b        2            +      3        ≥      18                  ∑                  c          y          c                                                  a            2                    +          a          b          +          2          a          +                      b            2                    +          3                                      a            2                    +          a          b          −          2          a          +                      b            2                    +          3                    +      3        .Now, you want to use the following reasoning.Since       ∑          c      y      c                          a        2            +      a      b      +      2      a      +              b        2            +      3                      a        2            +      a      b      −      2      a      +              b        2            +      3        =  3  +      ∑          c      y      c            (                            a          2                +        a        b        +        2        a        +                  b          2                +        3                              a          2                +        a        b        −        2        a        +                  b          2                +        3              −    1    )    =  =  3  +      ∑          c      y      c                  4      a                      a        2            +      a      b      −      2      a      +              b        2            +      3        ≥  3  ,where the equality occurs for a=b=c=0, we see that the minimal value of      ∑          c      y      c                          a        2            +      a      b      +      2      a      +              b        2            +      3                      a        2            +      a      b      −      2      a      +              b        2            +      3      is 3.Thus, it&#039;s enough to prove that      ∑          c      y      c                          a        2            +      a      b      −      2      a      +              b        2            +      3                      a        2            +      a      b      +              b        2            +      3        ≥      18          3      +      3      or      ∑          c      y      c                          a        2            +      a      b      −      2      a      +              b        2            +      3                      a        2            +      a      b      +              b        2            +      3        ≥  3and here you say that the starting inequality is proven.I think it&#039;s not so because the last inequality is wrong.Try   a  =  b  =  c  =  1.I hope it&#039;s clear now.

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