You should for the following function such that:

`f(x) = ln x - 2(x-1)/(x+1) `

The problem requests to prove that `f(x) >= 0 < = > x > 1`

You need to prove that the function reaches its minimum at x = 1, hence, you need to evaluate the zeroes of the equation `f'(x) = 0` such that:

`f'(x) = 1/x - (2(x+1)-2(x-1))/((x+1)^2)`

`f'(x) = 1/x - (2x+2-2x+2)/((x+1)^2)`

`f'(x) = 1/x - 4/((x+1)^2) => f'(x) = (x^2 + 2x + 1 - 4x)/(x(x+1)^2)`

`f'(x) = (x^2 -2x + 1)/(x(x+1)^2) =gt f'(x) = ((x-1)^2)/(x(x+1)^2) `

You need to solve `f'(x) = 0` such that:

`((x-1)^2)/(x(x+1)^2) = 0 =gt (x-1)^2 = 0 =gt x_1 = x_2 = 1`

Hence, since the zeroes of derivative are `x_1 = x_2 = 1` , then the function reaches its extreme at `x = 1` .

**Notice that derivative is positive for `xgt0` and it is negative for `xlt0` , hence, if `x in (1,oo),` the function `f(x) = ln x - 2(x-1)/(x+1)` is positive, thus `ln x - 2(x-1)/(x+1) gt= 0 =gt ln xgt= 2(x-1)/(x+1).` **