Solve in positive real numbers the system of equations: ( x 3 + y 3 ) ( y 3 + z 3 ) ( z 3 + x 3 ) = 8 , x 2 x + y + y 2 y + z + z 2 z + x = 3 2 .

Question

Solve in positive real numbers the system of equations:                    (                  x          3                +                  y          3                )        (                  y          3                +                  z          3                )        (                  z          3                +                  x          3                )        =        8        ,                                                  x            2                                x            +            y                          +                              y            2                                y            +            z                          +                              z            2                                z            +            x                          =                  3          2                .

verzaadtwr · Accepted Answer

Note that                                          y            2                                x            +            y                          +                              z            2                                y            +            z                          +                              x            2                                z            +            x                                              =                  (                                    x              2                                      x              +              y                                +          y          −          x          )                +                  (                                    y              2                                      y              +              z                                +          z          −          y          )                +                  (                                    z              2                                      z              +              x                                +          x          −          z          )                                                  =                              x            2                                x            +            y                          +                              y            2                                y            +            z                          +                              z            2                                z            +            x                          =                  3          2                        .            Therefore,                    x        2            +              y        2                    x      +      y        +                    y        2            +              z        2                    y      +      z        +                    z        2            +              x        2                    z      +      x        =  3    .By the AM-GM Inequality, we see that                    (*)                              (                                                    x                2                            +                              y                2                                                    x              +              y                                )                          (                                                    y                2                            +                              z                2                                                    y              +              z                                )                          (                                                    z                2                            +                              x                2                                                    z              +              x                                )                ≤        1                .            We claim that, for   a  ,  b  &amp;gt;  0,                    (#)                                                        a              3                        +                          b              3                                2                ≤                              (                                                            a                  2                                +                                  b                  2                                                            a                +                b                                      )                    3                        ,            and the inequality becomes an equality if and only if   a  =  b. To justify the claim, let   p  (  t  )  ∈  R  [  t  ] be the polynomial  p  (  t  )  :=  2              (              t        2            +      1      )        3    −  (  t  +  1      )    3    (      t    3    +  1  )    .Since   p  (  1  )  =  16  −  16  =  0, we know that   (  t  −  1  ) is a factor of   p  (  t  ). Now,      p    ′    (  t  )  =  12    t  (      t    2    +  1      )    2    −  3    (  t  +  1      )    2    (      t    3    +  1  )  −  3        t    2    (  t  +  1      )    3      ,which satisfies       p    ′    (  1  )  =  48  −  24  −  24  =  0 again. That is,   (  t  −  1      )    2   is a factor of   p  (  t  ). We proceed further:      p    ″    (  t  )  =  48        t    2    (      t    2    +  1  )  +  12    (      t    2    +  1      )    2    −  6    (  t  +  1  )  (      t    3    +  1  )  −  18        t    2    (  t  +  1      )    2    −  6    t  (  t  +  1      )    3      .We have again that       p    ″    (  1  )  =  96  +  48  −  24  −  72  −  48  =  0, and so   (  t  −  1      )    3   is a factor of   p  (  t  ). Now,      p    ‴    (  t  )  =  144    t  (      t    2    +  1  )  +  96        t    3    −  6    (      t    3    +  1  )  −  54        t    2    (  t  +  1  )  −  54    t  (  t  +  1      )    2    −  6    (  t  +  1      )    3      ,so       p    ‴    (  1  )  =  288  +  96  −  12  −  108  −  216  −  48  =  0, whence   (  t  −  1      )    4   is a factor of   p  (  t  ). Because   p is a monic polynomial of degree   6, we must have  p  (  t  )  =  (  t  −  1      )    4      (      t    2    +  α  t  +  β  )for some   α  ,  β  ∈  R. With   p  (  0  )  =  1, we get   β  =  1. As   p  (  −  1  )  =  16, we conclude that   1  −  α  +  β  =  1, so that   α  =  1, as well. Consequently,  p  (  t  )  =  (  t  −  1      )    4      (      t    2    +  t  +  1  )    ,which is a nonnegative polynomial (i.e.,   p  (  R  )  ⊆        R          ≥      0      ), and the only real root of   p  (  t  ) is   t  =  1.Now, (#) is equivalent to  2              (              a        2            +              b        2            )        3    −  (  a  +  b      )    3        (          a      3        +          b      3        )    =      b    6      p      (          a      b        )    =  (  a  −  b      )    4          (          a      2        +    a    b    +          b      2        )    ≥  0    ,which is an equality iff   a  =  b. Hence, the claim is established, but then we conclude that      (                            x          3                +                  y          3                    2        )        (                            y          3                +                  z          3                    2        )        (                            z          3                +                  x          3                    2        )    ≤  1    ,using (#) in (*). Therefore,  (      x    3    +      y    3    )  (      y    3    +      z    3    )  (      z    3    +      x    3    )  ≤  8    .However, the problem statement demands that the inequality above is an equality. That is,   x  =  y  =  z must hold. Ergo, the only positive real solution to this system of equations is  (  x  ,  y  ,  z  )  =  (  1  ,  1  ,  1  )    .

Solve in positive real numbers the system of equations: <mtable rowspacing="3pt" columnspacing=

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