Срочно! Помогите пожалуйста с неравенством

Для простоты:
(x^2 + 1) = a
(x - 3)= b
тогда уравнение примет вид:

log(a) b^2 * log(a) [b^2 \ a^3] =< -2

2 * log(a) b * [log(a) b^2 - log(a) a^3] =< -2

log(a) b^2 * [2 * log(a) b - 3 * log(a) a] =< -1

log(a) b * [2 * log(a) b - 3 * 1] =< -1

2 * {log(a) b}^2 - 3 * log(a) b =< -1 --------------------> log(a) b = t

2t^2 - 3t +1 =< 0

(t - 1) * (t - 2) =< 0 ---------------------------------> t1 < 1 или t2 > 2
=>
log(a) b = t1 ----------> log(a) b < 1
log(a) b = t2 ----------> log(a) b > 2
=>
log(x^2+1) [x - 3] < 1
lg (x^2+1) \ lg (x-3) < 1
lg (x^2+1) > lg (x-3) ---------> (x^2+1) < (x-3) -----> x^2 - x + 4 < 0 ----> D < 0 ----> решений нет

log(x^2+1) [x - 3] > 2
lg (x^2+1) \ lg (x-3) > 2
lg (x^2+1) > 2*lg (x-3)
lg (x^2+1) > lg (x-3)^2 ------> (x^2+1) > (x-3)^2 ----->
------> x^2 + 1 - x^2 + 6x - 9 > 0 ----------> 6x>8 ----------------> x > 4\3