# Another inequality to solve

Posted 4 months ago
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 Hi everyone, I'm trying to solve an equation that is similar to my previous post, but I don't get how it could be different and doesn't end running again. Do I forgot something? I'm putting the file with the code to be more practical. Thanks in advance for any help or leads!  Attachments: Answer
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Posted 4 months ago
 Welcome to Wolfram Community! Please make sure you know the rules: https://wolfr.am/READ-1STThe rules explain how to format your code properly. Formatting your code makes it easier for other members to copy your code and help you. You can still keep your notebook attached. Please EDIT your post and make sure code blocks start on a new paragraph and look framed and colored like this. int = Integrate[1/(x^3 - 1), x]; Map[Framed, int, Infinity] Alternatively you can use the button "Add Notebook" to view the contents of your attached notebook in your post. Answer
Posted 4 months ago
 In a fresh start of Mathematica this Condition34=Reduce[{(a-c β)^2(6 b^2+4 b λ+λ^2)/(8 b^3)>3(a-c α)^2/(4 b), a>c>0,1>p>0,1>t>0,1>b>0,0<λ<1,β>α>1,a>c α,a>c β},β] instantly gives me α<β Answer
Posted 4 months ago
 Actually I don't get how but I attached the wrong file ... This one is the good one ;) Attachments: Answer
Posted 4 months ago
 This is very slow Condition34=Reduce[{(-a+c β+t θ1)^2(6b^2+4b λ+λ^2)/(8b^3)> (3a^2-6a c α+3c^2 α^2+2a(-3+2p)t θ2+2c(3-2p)t α θ2+(3-4p+2p^2)t^2 θ2^2)/(4b), a>c>0,1>p>0,1>t>0,1>b>0,1>λ>0,β>α>1,a>c α,a>c β,0<θ1<θ2<θ3<1},β] Verify we can divide both sides Simplify[(6b^2+4b λ+λ^2)/(8b^3)>0, {a>c>0,1>p>0,1>t>0,1>b>0,1>λ>0,β>α>1,a>c α,a>c β,0<θ1<θ2<θ3<1}] returns True Try to make the problem easier, but this is still very slow Condition34=Reduce[{(-a+c β+t θ1)^2> (3a^2-6a c α+3c^2 α^2+2a(-3+2p)t θ2+2c(3-2p)t α θ2+(3-4p+2p^2)t^2 θ2^2)/(4b)/(6b^2+4b λ+λ^2)*(8b^3), a>c>0,1>p>0,1>t>0,1>b>0,1>λ>0,β>α>1,a>c α,a>c β,0<θ1<θ2<θ3<1},β] Try to make the problem easier Condition34=Reduce[{(-a+c β+t θ1)^2>q, a>c>0,1>p>0,1>t>0,1>b>0,1>λ>0,β>α>1,a>c α,a>c β,0<θ1<θ2<θ3<1},β]/. q->(3a^2-6a c α+3c^2 α^2+2a(-3+2p)t θ2+2c(3-2p)t α θ2+(3-4p+2p^2)t^2 θ2^2)/(4b)/(6b^2+4b λ+λ^2)*(8b^3) The output is very large because we have not applied any domain informationThis is still slow, but it may provide a more compact result for you Simplify[Condition34,{a>c>0,1>p>0,1>t>0,1>b>0,1>λ>0,β>α>1,a>c α,a>c β,0<θ1<θ2<θ3<1}] This (Reduce[{(-a+c β+t θ1)^2>q, a>c>0,1>p>0,1>t>0,1>b>0,1>λ>0,β>α>1,a>c α,a>c β,0<θ1<θ2<θ3<1},β]/. q->(3a^2-6a c α+3c^2 α^2+2a(-3+2p)t θ2+2c(3-2p)t α θ2+(3-4p+2p^2)t^2 θ2^2)/(4b)/(6b^2+4b λ+λ^2)*(8b^3))/. {Inequality[0, Less, c, Less, a]->True,Inequality[0, Less, p, Less, 1]->True, Inequality[0, Less, t, Less, 1]->True,Inequality[0, Less, b, Less, 1]->True, Inequality[0, Less, λ, Less, 1]->True,Inequality[0, Less, a, LessEqual, 1]->True, Inequality[θ1, Less, θ2, Less, 1]->True,Inequality[θ2, Less, θ3, Less, 1]->True, Inequality[α, Less, β, Less, a/c]->True,Inequality[0, Less, θ1, Less, 1]->True, Inequality[1, Less, α, Less, a/c]->True } `will be much faster than Simplify, but it will not make the result as simple and it is necessary to be very careful with the pattern matching to make certain the replacements are exactly correct. If you look at the result of this there will still be many inequalities that might be replaced with True and thus eliminated if it is possible to justify the replacements. The result is still very large, but it might be possible to use this method to get it to be small enough that Simplify could then be done in acceptable time.Please check all this very carefully. Answer