Re: How Do I Solve This System of 6 Inequalities?
- To: mathgroup at smc.vnet.net
- Subject: [mg30385] Re: How Do I Solve This System of 6 Inequalities?
- From: gl.cruciani at tin.it (Gianluca Cruciani)
- Date: Tue, 14 Aug 2001 03:45:37 -0400 (EDT)
- References: <firstname.lastname@example.org>
- Sender: owner-wri-mathgroup at wolfram.com
Andrzej Kozlowski <andrzej at tuins.ac.jp> wrote in message news:<9kg0av$i9a$1 at smc.vnet.net>... > > > 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 by a rational approaximation. > > ??? 1) from w-Sqrtu/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 variables. Sorry for the delay (vacations).