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Re: How Do I Solve This System of 6 Inequalities?

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  • Subject: [mg30385] Re: How Do I Solve This System of 6 Inequalities?
  • From: gl.cruciani at (Gianluca Cruciani)
  • Date: Tue, 14 Aug 2001 03:45:37 -0400 (EDT)
  • References: <9kg0av$i9a$>
  • Sender: owner-wri-mathgroup at

Andrzej Kozlowski <andrzej at> wrote in message news:<9kg0av$i9a$1 at>...
> One could already try to use one of the built in functions but to speed 
> up the solution further transformations are desirable. First of all, 
> since we are assuming all the variables are >0, we can replace the 
> condition  (z/v)(u - y) < 1/4 by the z(u-y)<v/4, the condition 
> (z*x/v) >   1 by z*x>v. There is a problem with the condition  
> (z*w/v)y/((z*w/v) - 1) - u < 0. This reduces to ((z*w)*y)/(w*z - v) - 
> u < 0. There are two cases here, w*z<v and w*z>v. In the first case we 
> get u*v-u*w*z+w*y*z<0 and in the second  u*v-u*w*z+w*y*z>0. So we shall 
> use two pairs of conditions instead of your original one. We could now 
> try to solve the problem using CylindricalAlgebraicDecomposition or 
> InequalitySolve but it will take a very long time. To do it faster we 
> need to replace Sqrt[15] by a rational approaximation.


1) from w-Sqrt[15]u/4<0 you have u>w;
2) from (z/v)(u-y)<1/4 you have u-y<v/(4z);
3) from ((zw/v)y/((zw/v)-1))-u<0 you have only u-y>uv/(wz), since from
the inequalities w-x>0 and zx/v>1 you get zw/v-1>0.

So, from 2),3) you have uv/(wz)<v/(4z) that means u<w/4, which is
clearly incompatible with 1), since we are dealing with real positive

Sorry for the delay (vacations).

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