Oups I just saw that my first and second relationals are the same. So I
have to solve for only 3 relationals.
I just went to try the wolfram alpha online calculator, I got :
1/ for the first relational :
solve((-k**8 + 8*k**6 - 8*k**4 - 16*k**2 - 64)/(k*(k**8 - 2*k**6 -
4*k**4 - 16*k**2 + 8)) > 0, k > 2, k <= 3)
alpha gives:
2<k<sqrt(2/3 (2+(53-3 sqrt(201))^(1/3)+(53+3 sqrt(201))^(1/3)))
2/ for the third one:
solve((k**8 - 6*k**6 + 8*k**2 + 96)/(k*(k**8 - 2*k**6 - 4*k**4 - 16*k**2 +
8)) >= 0, k > 2, k <= 3)
alpha gives:
sqrt(2/3+1/3 (404-36 sqrt(109))^(1/3)+1/3 2^(2/3) (101+9
sqrt(109))^(1/3))<=k<=3
3/ and for the fourth one:
solve((k**8 - 2*k**7 + 4*k**6 + 4*k**5 - 16*k**4 + 8*k**3 - 40*k**2 + 24*k
- 64)/(k*(k**8 - 2*k**6 - 4*k**4 - 16*k**2 + 8)) >= 0, k > 2, k <= 3)
alpha gives:
2<k<=3
Bruno
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