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I am trying to verify the following inequality Mathematica. This I wrote the fillowing code:

Assumptions = {0 < a < 1, 0 < b < 1, a^2 + b^2 == 1, 
   0 < x < 1/2, 0 < y < 1/2};
(a^{2 x - 1} (1 - a^2)^{(2 y - 1)/2} - 1) > 0 // Simplify
(* {-1 + a^(-1 + 2 x) (1 - a^2)^(-(1/2) + y)} > 0 *)

I am getting above output. I am unable to decide. Can anyone help.

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    $\begingroup$ You will need to change $Assumptions which is the global definition - not Assumptions. $\endgroup$ – gwr Jul 8 '16 at 7:37
  • $\begingroup$ I do not know if you would accept it as an answer but $\endgroup$ – cyrille.piatecki Jul 8 '16 at 17:29
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Maybe:

expression = a^(2 x - 1) (1 - a^2)^((2 y - 1)/2 - 1);
assumptions = 0 < a < 1 && 0 < b < 1 && a^2 + b^2 == 1 && 0 < x < 1/2 && 0 < y < 1/2;

Positive[expression]~Refine~(assumptions)

(*True*)
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First of all your code is false

Assuming[0 < a < 1 && a^2 <= 1 && 0 < x < 1/2 && 0 < y < 1/2, 
Positive[a^(2 x - 1) (1 - a^2)^((2 y - 1)/2 - 1)]] // Simplify

I have put aside b since it does not appear in the expression to test. This does not work. In revanche,

Assuming[0 < a < 1 && a^2 <= 1 && 0 < x < 1/2 && 0 < y < 1/2, 
Element[a^(2 x - 1) (1 - a^2)^((2 y - 1)/2 - 1), Reals]] // Simplify 

Return positive. As $0< a < 1$, it should be the (positive) answer. Now you can also verify by looking at contour levels

F[x_, y_, a_] := a^(-1 + 2 x) (1 - a^2)^(-(3/2) + y)
ContourPlot[F[x, y, .3], {x, 0, 1/2}, {y, 0, 1/2}, 
PlotLegends -> Automatic]

in varying a

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  • $\begingroup$ of course its is a way to check. The reason why Positive does not work is that it neads numeric values. $\endgroup$ – cyrille.piatecki Jul 8 '16 at 19:59

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