1/(4^x - 1) - 2/(4^x + 1) < 1/(16^x - 1)
1/(4^x - 1) - 2/(4^x + 1) - 1/(16^x - 1) < 0
16^x - 1 = (4^x - 1)·(4^x + 1)
(1·(4^x + 1) - 2·(4^x - 1) - 1)/(16^x - 1) < 0
((2^(2·x) + 1) - (2^(2·x + 1) - 2) - 1)/(16^x - 1) < 0
(2 - 2^(2·x))/(16^x - 1) < 0
(2 - 2^(2·x))/(2^(4·x) - 1) < 0
2^(2·x) = t >0
(2 - t)/(t^2 - 1) < 0
risolvo ed ottengo: -1 < t < 1 ∨ t > 2
0 < 2^(2·x) < 1 ∨ 2^(2·x) > 2
x < 0 ∨ x > 1/2