Find the smallest number α, such that for all x , y , z α ( x 2 − x + 1 ) ( y 2 − y + 1 ) ( z 2 − z + 1 ) ≥ ( x y z ) 2 − x y z + 1My work so far:Let f ( t ) = t 2 − t + 1. Then f ( t ) ≥ 3 4 If x = 0 , y = z = 1 2 , then α ≥ 16 9

Question

Find the smallest number   α, such that for all   x  ,  y  ,  z  α  (      x    2    −  x  +  1  )  (      y    2    −  y  +  1  )  (      z    2    −  z  +  1  )  ≥  (  x  y  z      )    2    −  x  y  z  +  1My work so far:Let   f  (  t  )  =      t    2    −  t  +  1. Then   f  (  t  )  ≥      3    4  If   x  =  0  ,  y  =  z  =      1    2  , then  α  ≥      16    9

popman14ee · Accepted Answer

In the starting formulation the answer is       16    9  If   x,   y and   z are non-positives so after replacing   x  →  −  x... we need to prove that      16    9    (      x    2    +  x  +  1  )  (      y    2    +  y  +  1  )  (      z    2    +  z  +  1  )  ≥      x    2        y    2        z    2    −  x  y  z  +  1which is true for all non-negatives   x,   y and   z because we&#039;ll prove now that even      16    9    (      x    2    +  x  +  1  )  (      y    2    −  y  +  1  )  (      z    2    −  z  +  1  )  ≥      x    2        y    2        z    2    +  x  y  z  +  1If   x  ≤  0,   y  ≤  0 and   z  ≥  0 we need to prove that      16    9    (      x    2    +  x  +  1  )  (      y    2    +  y  +  1  )  (      z    2    −  z  +  1  )  ≥      x    2        y    2        z    2    −  x  y  z  +  1for non-negatives   x,   y and   z, which follows from      16    9    (      x    2    +  x  +  1  )  (      y    2    −  y  +  1  )  (      z    2    −  z  +  1  )  ≥      x    2        y    2        z    2    +  x  y  z  +  1again.Now we&#039;ll prove it:                          16        (                  x          2                +        x        +        1        )        (                  y          2                −        y        +        1        )        (                  z          2                −        z        +        1        )        −        9        (                  x          2                          y          2                          z          2                +        x        y        z        +        1        )                                                        a            a            a                          =        (        16        (                  y          2                −        y        +        1        )        (                  z          2                −        z        +        1        )        −        9                  y          2                          z          2                )                  x          2                +        (        16        (                  y          2                −        y        +        1        )        (                  z          2                −        z        +        1        )        −        9        y        z        )        x                                                        a            a            a            a            a                          +        16        (                  y          2                −        y        +        1        )        (                  z          2                −        z        +        1        )        −        9                                                        a            a            a                          ≥                  (          16                      (                          3              4                                      y              2                        +                                          (                                  y                  2                                −                1                )                            2                        )                                (                          3              4                                      z              2                        +                                          (                                  z                  2                                −                1                )                            2                        )                    −          9                      y            2                                z            2                    )                          x          2                +        +        (        3        y        ⋅        3        z        −        9        y        z        )        x                                                        a            a            a            a            a                          +        16                  (                      3            4                    +                                    (                              1                2                            −              y              )                        2                    )                          (                      3            4                    +                                    (                              1                2                            −              z              )                        2                    )                −        9                                                        a            a            a                          ≥        0

Find the smallest number α , such that for all x , y , z

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