Prove the inequality ln(x-2)>=2(x-3)/(x-1), x>=3.

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).