Available after finishing, e x- 1/2 x 2-a x > = 1, because when x >; = 1/2, where e x is increasing function, that is, e x >；; = e 1/2,-1/2x 2 is a decreasing function, that is,-1/2x2; = 1/2，e x- 1/2 x2 > =e^ 1/2- 1/8

So ax < = e1/2-9/8 (x >; = 1/2)

When a=0, when x >; = 1/2，f(x)>； = 5/2 x 2+(a-3) x+ 1 hold.

When a<0 and X> are equal to1/2, ax is a decreasing function, and the maximum value of ax is 1/2a, f (x) >; = 5/2 x 2+(a-3) x+ 1 holds, so a

When a>0, when X>= 1/2, ax is increasing function, the minimum value of ax is 1/2a, and f(x) > = 5/2 x 2+(a-3) x+ 1 holds.

A<= e 1/2-9/8, and because 2e 1/2-9/4 is greater than 0, a > 2e1/2-9/4.

So the range of a is [0, -∞) u (2e 1/2-9/4, ∞).