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Reduce[{(4 I1^2 (-2 + b (-2 + I1)^2 + 2 I1 - I1^2) + 
  c^2 (-3 - I1^4 + b (-3 + I1^2)^2) - 
  4 c I1 (-2 + I1^2 - I1^3 + b (6 - 3 I1 - 2 I1^2 + I1^3))) ((
 c (-1 + I1) (c - 2 I1 + c I1))/(b I1^2 (a - r)))^b < 0, c > 0, 
 c < 1, b > 1, r > a, r > 0, a > 0}, I1, Reals]

The code above cannot be "reduced" in Mathematica. Could someone tell me what I am doing wrong?

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it seems that your first term is a product $ A B^b<0 $ with $ A=c^2 \left(b \left(\text{I1}^2-3\right)^2-\text{I1}^4-3\right)-4 c \text{I1} \left(b \left(\text{I1}^3-2 \text{I1}^2-3 \text{I1}+6\right)-\text{I1}^3+\text{I1}^2-2\right)+4 \text{I1}^2 \left(b (\text{I1}-2)^2-\text{I1}^2+2 \text{I1}-2\right), $ $ B=\left(\frac{c (\text{I1}-1) (c \text{I1}+c-2 \text{I1})}{b \text{I1}^2 (a-r)}\right)^b $, $ A,B \neq 0, $ so why not formulate the problem into two: $ A<0, B>0 $ and $ A>0, B<0, n\; \text{odd}. $ –  gpap Sep 20 '13 at 11:09
    
even if I do that I still have the same reduction problem. –  bubbly Sep 20 '13 at 11:24
1  
I don't think you do. Try With[{A = (4 I1^2 (-2 + b (-2 + I1)^2 + 2 I1 - I1^2) + c^2 (-3 - I1^4 + b (-3 + I1^2)^2) - 4 c I1 (-2 + I1^2 - I1^3 + b (6 - 3 I1 - 2 I1^2 + I1^3))), B = ((c (-1 + I1) (c - 2 I1 + c I1))/(b I1^2 (a - r)))}, Reduce[{A < 0, B > 0, 0 < c < 1, b > 1, r > a, r > 0, a > 0}, I1, Reals]] (valid for any $ b>1$), and replace A > 0, B < 0 which will be valid for only odd bs. –  gpap Sep 20 '13 at 11:30
    
Yes, that works ! Thanks! –  bubbly Sep 20 '13 at 16:17
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