# Hello!:) How can I solve this inequality?

GROUPS:
 Hello everybodyMathematica is really great! I am new here and I kindly wanted to ask you a question about solving an inequality with parameters.So, I firstly solved the following expression (with all the parameters >0):FullSimplify [ Reduce [ (Gu  A  h -  A  Gr - P  h +  r) / (2 b N + N T) > (Gu  A  h  -  A  Gr  -  P h) / (2 b N)  && h > 0 && b > 0 &&T > 0 && r > 0 && N > 0 && Gu > 0 && A > 0 && Gr > 0 ]]which gave me as a result:T > 0 && r > 0 && h > 0 && Gu > 0 && Gr > 0 && b > 0 && A > 0 && N > 0 && A Gu h T < 2 b r + A Gr T + h P T      (and this was ok)The problem is that now, I wanted to introduce an additional parameter (D<1)Rewriting the expression:FullSimplify [ Reduce [ (Gu  A  h^D -  A  Gr - P  h +  r) / (2 b N + N T) > (Gu  A  h^D  -  A  Gr  -  P h) / (2 b N) && h > 0 && b > 0 &&T > 0 && r > 0 && N > 0 && Gu > 0 && A > 0 && Gr > 0 && D<1 ]]I get:" Reduce::nsmet: This system cannot be solved with the methods available to Reduce. " and I also get as Out:Reduce[b N (2 b + T) (2 b r + (A (Gr - Gu h^D) + h P) T) > 0 &&h > 0 && b > 0 && T > 0 && r > 0 && N > 0 && Gu > 0 && A > 0 && Gr > 0 && D < 1]So my question is:  if I cannot solve the system with this new parameter D<1 (but why????) what is this result?I apologize if this question looks stupid but it is giving me a lot of problems and I don't know how to solve it!Thank you very much, in the case someone would like to help me!Kodi
4 years ago
8 Replies
 In some situations, Reduce works better if it is used step-wise.  Since your first Reduce calculation works, I would try Reduce on the result of the first calculation and the added constraint.
4 years ago
 Hi Frank!Thank you very much for your kind replySorry..I am little bit confused..you're saying to use Reduce on the first correct result (belonging to the first correct inequality), adding the condition D<1? But how can I do that? I mean, D is not contained in the first inequality..:/Sorry if I understood wrong, I'm a little bit confused..
4 years ago
 Frank Kampas 1 Vote Which version of Mathematica are you using?  If I runReduce[(Gu A h^D - A Gr - P h + r)/(2 b N + N T) > (Gu A h^D - A Gr -       P h)/(2 b N) && h > 0 && b > 0 && T > 0 && r > 0 && N > 0 &&   Gu > 0 && A > 0 && Gr > 0 && D < 1]in Version 9, i geth^D \[Element]   Reals && ((h^(-1 + D) \[Element] Reals && A > 0 && b > 0 && D < 1 &&      Gr > 0 && Gu > 0 && h > 0 && N > 0 && P < (-A Gr + A Gu h^D)/h &&      r > 0 && 0 < T < -((2 b r)/(A Gr - A Gu h^D + h P))) || (A > 0 &&      b > 0 && D < 1 && Gr > 0 && Gu > 0 && h > 0 && N > 0 &&      P >= (-A Gr + A Gu h^D)/h && r > 0 && T > 0))
4 years ago
 Dear Frank,thank you very much!I am also using the version 9.I apologize, I made a mistake, I forgot the condition P>0; if I clarify that condition, I can't get a result! Moreover, in trying the calculations without P>0, the parameter I had introduced was originally not labled "D" but "alpha" (I didn't think that there could be a difference!:/  I just used D in this forum as I couldn't write alpha). I noticed that in doing the same procedure, with "D" I get your same result, while with "alpha" I cannot. How is this possible?..I saw that alpha appears labeled in blue, while D in black..is there a meaning??Still, if I use "E" I obtain "False" ???Why does the result change in changing the letter ???Anyways the problem now is also due to the fact that if I insert P>0 (which I had forgotten) I can't get a result anymore, neither with D!
4 years ago
 Ilian Gachevski 1 Vote D (ref) and E (ref) and N (ref) are probably not the best choices for variable names.