Prove that if x + y 3 a − b = y + z 3 b − c = z + x 3 c − a then x + y + z a + b + c = a x + b y + c z a 2 + b 2 + c 2 I tried to prove this in many ways. First, I tried to multiply the first equality by ( 3 a − b ) ( 3 b − c ) ( 3 c − a ) , but then it seems too complicated. Is there any easy method?

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Prove that if              x      +      y              3      a      −      b        =            y      +      z              3      b      −      c        =            z      +      x              3      c      −      a       then             x      +      y      +      z              a      +      b      +      c        =            a      x      +      b      y      +      c      z                      a        2            +              b        2            +              c        2            I tried to prove this in many ways. First, I tried to multiply the first equality by       (    3    a    −    b    )    (    3    b    −    c    )    (    3    c    −    a    )   , but then it seems too complicated. Is there any easy method?

Aryanna Caldwell · Accepted Answer

If we set   k  =            x      +      y              3      a      −      b        =            y      +      z              3      b      −      c        =            z      +      x              3      c      −      a        ,then we have  x  +  y  =  (  3  a  −  b  )  k  ,        y  +  z  =  (  3  b  −  c  )  k  ,        z  +  x  =  (  3  c  −  a  )  k  .Hence, adding these three gives us                    (1)                    2        (        x        +        y        +        z        )        =        (        3        a        −        a        +        3        b        −        b        +        3        c        −        c        )        k                ⟺                                      x            +            y            +            z                                a            +            b            +            c                          =        k            On the other hand, we have  a  (  x  +  y  )  =  a  (  3  a  −  b  )  k  ,        b  (  y  +  z  )  =  b  (  3  b  −  c  )  k  ,        c  (  z  +  x  )  =  c  (  3  c  −  a  )  k                    (2)                    ⇒        (        a        x        +        b        y        +        c        z        )        +        (        a        y        +        b        z        +        c        x        )        =        {        3        (                  a          2                +                  b          2                +                  c          2                )        −        (        a        b        +        b        c        +        c        a        )        }        k              b  (  x  +  y  )  =  b  (  3  a  −  b  )  k  ,        c  (  y  +  z  )  =  c  (  3  b  −  c  )  k  ,        a  (  z  +  x  )  =  a  (  3  c  −  a  )  k                    (3)                    ⇒        (        a        x        +        b        y        +        c        z        )        +        (        a        z        +        b        x        +        c        y        )        =        {        3        (        a        b        +        b        c        +        c        a        )        −        (                  a          2                +                  b          2                +                  c          2                )        }        k              c  (  x  +  y  )  =  c  (  3  a  −  b  )  k  ,        a  (  y  +  z  )  =  a  (  3  b  −  c  )  k  ,        b  (  z  +  x  )  =  b  (  3  c  −  a  )  k                    (4)                    ⇒        (        a        y        +        b        z        +        c        x        )        +        (        a        z        +        b        x        +        c        y        )        =        {        3        (        a        b        +        b        c        +        c        a        )        −        (        a        b        +        b        c        +        c        a        )        }        k            Hence, calculating   (  2  )  +  (  3  )  −  (  4  ) gives us                    (5)                    2        (        a        x        +        b        y        +        c        z        )        =        2        (                  a          2                +                  b          2                +                  c          2                )        k                ⟺                                      a            x            +            b            y            +            c            z                                              a              2                        +                          b              2                        +                          c              2                                      =        k            From (1) and (5), we have the conclusion.

Prove that if x + y </mrow> 3 a &#x

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