Trying to solve a system of inequalitiesI have this kind of system of inequalities (with one equality):1. x + y + z = 12. a x + b y + c z ≤ 23. a 2 x + b 2 y + c 2 z ≤ 64. a 3 x + b 3 y + c 3 z ≤ 14and so on.. (I could continue with 5. 6. 7.... the power always increases by one and I know the value on the right).x,y,z are my unknowns and a,b,c are known.Could I get a solution (or an approximation) for x,y and z as the number of inequalities grows, I don't see it?

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Trying to solve a system of inequalitiesI have this kind of system of inequalities (with one equality):1.   x  +  y  +  z  =  12.   a  x  +  b  y  +  c  z  ≤  23.       a    2    x  +      b    2    y  +      c    2    z  ≤  64.       a    3    x  +      b    3    y  +      c    3    z  ≤  14and so on.. (I could continue with 5. 6. 7.... the power always increases by one and I know the value on the right).x,y,z are my unknowns and a,b,c are known.Could I get a solution (or an approximation) for x,y and z as the number of inequalities grows, I don&#039;t see it?

Danube2w · Accepted Answer

Not a full answer, but too long for a comment: let&#039;s first assume that all quantities involved are positive for convenience. Let   u  =  (  x  ,  y  ,  z  ), and let       v    n    =  (      a    n    ,      b    n    ,      c    n    ). The inequalities translate to the system      {                            u          ⋅                              1            →                    =          1                                      u          ⋅                      v            k                    ≤                      b            k                    ,                    1          ≤          k          ≤          n                        for given constants       b    k  . By Cauchy-Schwarz, we have that   u  ⋅      v    k    ≤  |  u  |  |      v    k    |,   u  ⋅      v    k    ≤  |  u  |  |      v    k    | ,  so a sufficient (but certainly not necessary) condition for the system of inequalities to hold is   |  u  |  |      v    k    |  ≤      b    k  for each   k, equivalently      x    2    +      y    2    +      z    2    ≤            b      k      2                      a                  2          k                    +              b                  2          k                    +              c                  2          k                      .Now, we run through   1  ≤  k  ≤  n and calculate  m  =      min          1      ≤      k      ≤      n            {                  b        k        2                              a                      2            k                          +                  b                      2            k                          +                  c                      2            k                                }    .Once we calculate this, then we are left with the restriction   |  u  |  &amp;lt;      m   which is an open ball in         R    3  . On the other hand   u  ⋅      1    →    =  1 represents a plane, so taking any vector   u which lies on their intersection is sufficient.This is not a full solution, however: the method can fail because there might not be any   u in their intersection, although there does exist   u satisfying the original system. This is because of the use of Cauchy-Schwarz, which is not an if-and-only-if criterion.

Trying to solve a system of inequalities I have this kind of system of inequalities (with one equal

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