So let g (x) = e x-x- 1, then g' (x) = e x- 1, if x >；; =0，g'(x)>=0，0 & gt； x & gt- 1，g' (x) < 0

Therefore, the minimum value of x & gtg(x) at-1 is g(0)=0, so g(x) is at x >; When =- 1, it is greater than or equal to 0, so it is proved.

2。 When x & gt=0, because f(x) is obviously greater than or equal to 0 and less than 1, when A < 0, if X/(AX+ 1) < 0, 0 < x <-1/a, so when x

So a≥0, then let h(x)=f(x)(ax+ 1)-x, as long as h(x)

When a> is at 1/2, there is h (x) >: = (a-1-ax) * x/(x+1)+ax = (2a-1) x/(x+/). 0

So to sum up, 0