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.

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.