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We'll create the helping function f(x) = ln(x-2) - 2(x-3)/(x-1), that is differentiable over the domain of the function.
We'll differentiate and we'll get:
f'(x) = 1/(x-2) - 4/(x-1)^2
f'(x) = [(x-1)^2 - 4x + 8]/(x-2)*(x-1)^2
f'(x) = (x^2 - 2x + 1 - 4x - 8)/(x-2)*(x-1)^2
f'(x) = (x-3)^2/(x-2)*(x-1)^2
We notice that for any value of x, the derivative will be positive, since it's numerator and denominator are always positive.
We also notice that f'(3) = 0
Then the function is increasing over [3 ; +infinite).
For x>=3 => f(x)>=f(3) = 0
If f(x) > 0 => ln(x-2) - 2(x-3)/(x-1)>0
The inequality is true for any value of x>=3: ln(x-2) >=2(x-3)/(x-1).
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