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Obtain (faster) conditions from inequalities?

Posted 8 years ago

This is not a homework, so please do not erase my question. Thank you.

I am interested in obtaining conditions for $a,b,c,d > 0 \quad \forall \epsilon >0 $ from four inequalities.

$$ w_1 < 1 + w_2 + w_3, $$ $$1 < w_1 + w_2 + w_3,$$ $$w_2 < 1 + w_1 + w_3,$$ $$ w_3 < 1 + w_2 + w_1$$

where

$$w_1 = \sqrt{\frac{a(1 + d\epsilon)}{((ad - bc)\epsilon+ a + d - b - c)}},$$ $$ w_2 = \sqrt{\frac{d(1 + a\epsilon)}{((ad - bc)\epsilon+ a + d - b - c)}},$$ $$w_3 = \sqrt{\frac{(ad-bc)\epsilon}{((ad - bc)\epsilon+ a + d - b - c)}}$$

w1 = Sqrt[(a*(1 + d*\[Epsilon]))/((a*d - b*c)*\[Epsilon] + a + d - b -c)],   
w2 = Sqrt[(d*(1 + a*\[Epsilon]))/((a*d - b*c)*\[Epsilon] + a + d - b - c)]   
w3 = Sqrt[((a*d - b*c)*\[Epsilon])/((a*d - b*c)*\[Epsilon] + a + d - b - c)]      

Furthermore, I calculated already some restrictions for a,b,c,d.

I used

Reduce[{a > 0, b > 0, c > 0, d > 0, \[Epsilon] > 0, a > d , a*d > b*c, a + d >= b + c, w1 < 1 + w2 + w3, 1 < w1 + w2 + w3, 
  w2 < 1 + w1 + w3, w3 < 1 + w2 + w1}, {a, b, c, d}, Reals]

Mathematica is running now for hours and I am wondering if there is a problem with the command or is it just impossible to get some conditions for a,b,c,d?

I hope, someone can help me. Thank you.

POSTED BY: P D

Please EDIT your post and format code properly according to this tutorial: http://wolfr.am/READ-1ST

Not properly formatted code can become corrupted in editor and give errors.

POSTED BY: Moderation Team
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