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Given inequality f[x, y]>0 where f[x, y]=1/16 (-1 + x (2 - x + x^3 (-1 + y)^2 y^2)), how can one find the values x (keeping y fixed) such that the inequality is satisfied. And then repeat the same to find y (keeping x fixed)? The answer appears in equation (107) of this article:

$$2(\sqrt{2}-1) \le x \le 1, \qquad \frac{1}{2}\left( 1- \sqrt{ \frac{x^2 +4x - 4}{x^2} } \right) \le y \le \frac{1}{2}\left( 1+\sqrt{ \frac{x^2 +4x - 4}{x^2} } \right)$$

Edit: The parameters $x$ and $y$ satisfy $0\le x \le 1$ and $0 \le y \le 1$.

A similar but related problem with four variables $0\le x1 \le 1$, $0\le y1 \le 1$, $0\le x2 \le 1$, $0\le y2 \le 1$, and this time

 f[x1_, y1_, x2_, y2_]=1/128 (2 + x1 (-2 + x2) y1 + (-2 + x1) x2 y2) (1 + x1 (-1 + y1) + 
   x2 (-1 + y2) - 
   1/2 x1 x2 (-2 + y1 + y2)) (-4 (-1 + x1) (-1 + 
      x2) - (x1 (-2 + x2) y1 + (-2 + x1) x2 y2) (-2 x2 (-1 + y2) + 
      x1 (2 - 2 y1 + x2 (-2 + y1 + y2))))
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  • $\begingroup$ I recommend editing the question with a brief explanation what x and y stand for and what the inequality means. Moreover the title should be restricted to Information-theoretic or/and quantum compiting ideas. Then the question would be more interesting for readers searching for related topics. $\endgroup$
    – Artes
    Commented Aug 24, 2021 at 23:43

1 Answer 1

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It seems the problem is not precisely formulated. Nonetheless we can reconstruct reasoning behind the result obtained in the article metioned.

Let's define

f[x_, y_] := 1/16 (-1 + x (2 - x + x^3 (-1 + y)^2 y^2))

then it is straightforward to exploit Reduce this way:

Reduce[ f[x, y] > 0, y]

nevertheless the result is much more involved than the formula (107). So we deduce that the both 0 <= x <= 1 and 0 <= y <= 1 should be satisfied as well, and so we get

Reduce[f[x, y] > 0 && 0 <= x <= 1 && 0 <= y <= 1, y]
  2 (-1 + Sqrt[2]) < x <= 1 && 
 1/2 - 1/2 Sqrt[(-4 + 4 x + x^2)/x^2] < y < 1/2 + 1/2 Sqrt[(-4 + 4 x + x^2)/x^2]

analogically we can obtain the result assuming that y is fixed by

Reduce[f[x, y] > 0 && 0 <= x <= 1 && 0 <= y <= 1, x]
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  • $\begingroup$ thanks a lot. I have a similar problem, but this time with 4 parameters (as given in the edited section). Do you see any hope of finding similar "ranges" for x1, y1, x2, y2? $\endgroup$
    – Zubin
    Commented Aug 25, 2021 at 11:07

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