Prove that ( 2 a + b + c ) 2 2 a 2 + ( b + c ) 2 + ( 2 b + c + a ) 2 2 b 2 + ( c + a ) 2 + ( 2 c + a + b ) 2 2 c 2 + ( a + b ) 2 ≤ 8Prove that ( 2 a + b + c ) 2 2 a 2 + ( b + c ) 2 + ( 2 b + c + a ) 2 2 b 2 + ( c + a ) 2 + ( 2 c + a + b ) 2 c 2 + ( a + b ) 2 ≤ 8MY ATTEMPT:I want to make a relation between a , b , c. By trial I found that if we put a = b = c = 1 then the above inequality holds(equality also holds). So by trial I assume that a + b + c = 3. After that the three functions become of the form of the function below: f ( x ) = ( x + 3 ) 2 2 x 2 + ( 3 − x ) 2 . I calculate the function and found that : f ( x ) ≤ ⅓ ( 4 x + 4 ). Am I do right . Anybody has other ideas.

Question

Prove that             (      2      a      +      b      +      c              )        2                    2              a        2            +      (      b      +      c              )        2              +            (      2      b      +      c      +      a              )        2                    2              b        2            +      (      c      +      a              )        2              +            (      2      c      +      a      +      b              )        2                    2              c        2            +      (      a      +      b              )        2              ≤  8Prove that            (      2      a      +      b      +      c              )        2                    2              a        2            +      (      b      +      c              )        2              +            (      2      b      +      c      +      a              )        2                    2              b        2            +      (      c      +      a              )        2              +            (      2      c      +      a      +      b      )              2              c        2            +      (      a      +      b              )        2              ≤  8MY ATTEMPT:I want to make a relation between   a  ,  b  ,  c. By trial I found that if we put   a  =  b  =  c  =  1 then the above inequality holds(equality also holds). So by trial I assume that   a  +  b  +  c  =  3. After that the three functions become of the form of the function below:   f  (  x  )  =            (      x      +      3              )        2                    2              x        2            +      (      3      −      x              )        2            . I calculate the function and found that :   f  (  x  )  ≤      ⅓    (  4  x  +  4  ). Am I do right . Anybody has other ideas.

kamphundg4 · Accepted Answer

For non-negatives   a,   b and   c let   a  +  b  +  c  =  3. Hence,  8  −      ∑          c      y      c                  (      2      a      +      b      +      c              )        2                    2              a        2            +      (      b      +      c              )        2              =      ∑          c      y      c            (          8      3        −                  (        a        +        3                  )          2                            2                  a          2                +        (        3        −        a                  )          2                      )    =  =      1    3        ∑          c      y      c                  (      a      −      1      )      (      7      a      −      15      )                      a        2            −      2      a      +      3        =      1    3        (          ∑              c        y        c                            (        a        −        1        )        (        7        a        −        15        )                              a          2                −        2        a        +        3              +    4    (    a    −    1    )    )    =  =      ∑          c      y      c                  (      a      −      1              )        2            (      4      a      +      3      )              3      (              a        2            −      2      a      +      3      )        ≥  0Also there is the following.It&#039;s enough to prove that            (      2      a      +      b      +      c              )        2                    2              a        2            +      (      b      +      c              )        2              ≤            4      (      4      a      +      b      +      c      )              3      (      a      +      b      +      c      )        ,which is   (  2  a  −  b  −  c      )    2    (  5  a  +  b  +  c  )  ≥  0

Prove that ((2a+b+c)^2)/(2a^2+(b+c)^2)+((2b+c+a)^2)/(2b^2+(c+a)^2)+((2c+a+b))/(2c^2+(a+b)^2)<=8

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