(2)

Multiply f' (x) =1/x-a.

F(x) is in (0, 1/a) ↗ and (1/a, +∞) ↘.

∴f(x)max=ln( 1/a)-a*( 1/a)+4a？ =-lna+4a？ - 1

Let g(a)=f(x)max.

∴g'(a)=- 1/a+8a

G(a) in (0, √2/4)↗) and in (√2/4, +∞) ↘.

Therefore, when the maximum value M(a) of f(x) exists, a=√2/4.

∫f "(x)=-( 1/x)？ & lt0

∴f'(x) in (0, +∞) ↘

As can be seen from the f'(x) diagram, the change of f'(x) is intense at first, and then slowly slows down.

The f(x) image thus obtained looks like this:

∴a 1+a2>； 2a=√2/2

Namely a1+a2 >; √2/2