Then, f (x) >; G(x) holds forever.

x^2-ax>； B+aln(x- 1) gives b.

Let f (x) = x 2-ax-AlN (x-1).

Get f' (x) = 2x-a-a/(x- 1)。

The minimum value of F(x) is f (1+a/2) =1-a * a-alna/2.

So b < 1-a*a-alna/2 holds.

While 1-a*a-alna/2 is decreasing. It reaches the maximum when a= 1, and the maximum value is ln2.

So b < ln2