How to solve the following system of inequalities rigorously { Φ ≥ 1 2 ( x 2 − y 2 − z 2 ) + x + y + z − 3 8 , Φ ≤ 1 2 ( x 2 − y 2 − z 2 ) + 1 2 x + y + z − 9 8 Φ ≤ 1 2 ( x 2 − y 2 − z 2 ) + 1 8 y ≥ 1 2 , z ≥ 1 The task is find the range of Φ such that this system of inequalities in x , y , z has solutions in R.

Question

How to solve the following system of inequalities rigorously      {                            Φ          ≥                      1            2                    (                      x            2                    −                      y            2                    −                      z            2                    )          +          x          +          y          +          z          −                      3            8                    ,                                      Φ          ≤                      1            2                    (                      x            2                    −                      y            2                    −                      z            2                    )          +                      1            2                    x          +          y          +          z          −                      9            8                                                Φ          ≤                      1            2                    (                      x            2                    −                      y            2                    −                      z            2                    )          +                      1            8                                                y          ≥                      1            2                    ,          z          ≥          1                        The task is find the range of   Φ such that this system of inequalities in   x  ,  y  ,  z has solutions in   R.

Samantha Reid · Accepted Answer

There is no upper limit for   Φ, as   x can be chosen arbitrarily high to fulfil the second and the third inequality.The first inequality can be rewritten to:  2  Φ  ≥  (      x    2    +  2  x  +  1  )  −  (      y    2    −  2  y  +  1  )  −  (      z    2    −  2  z  +  1  )  +      1    4  Inserting the inequalities for   y and   z:  2  Φ  ≥  (  x  +  1      )    2  Therefore,   Φ must be larger than or equal to zero.

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