S= 1/2ab*sinC，

So a 2+b 2-c 2 = 2ab * (1-1/4 * sinc),

So cosc = (A2+B2-C2)/2ab =1-1/4 * sinc,

1/4*sinC+cosC= 1，

√ 17/4*sin(C+t)= 1, (where cost=√ 17/ 17, sint=4√ 17/ 17).

So sin (c+t) = 4 √17/17 = sin (π-t),

C+t=π-t，

C=π-2t .

So the angle c is a constant.

And a+b=2, and both a and b are positive numbers. From the average inequality, we get:

√(ab)& lt； =(a+b)/2= 1，

If and only if a=b= 1, take the equal sign,

So ab < = 1,

So s = 1/2ab * sinc.

Therefore, the maximum value of the area s of triangle ABC is 4/ 17.