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Another inequality to solve

Posted 1 month 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! enter image description here

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Welcome to Wolfram Community! Please make sure you know the rules: https://wolfr.am/READ-1ST

The 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]

enter image description here

Alternatively you can use the button "Add Notebook" to view the contents of your attached notebook in your post.

Posted 1 month 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<a/c-Sqrt[6]*Sqrt[(a^2*b^2-2*a*b^2*c*α+b^2*c^2*α^2)/(c^2*(6*b^2+4*b*λ+λ^2))]]

I have no evidence that it might be the cause of the problem, but you might remember that if you assign a value to a variable then Mathematica will silently remember that assignment until you clear that or restart Mathematica. Some users are surprised when hours or days later something doesn't work and it is finally discovered that they didn't realize a previous assigned value was being used.

Please check all this very carefully.

Actually I don't get how but I attached the wrong file ... This one is the good one ;)

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Posted 1 month 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 information

This 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.

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