not sure if this is a valid proof, but I'll give it a shot :-)

without loss of generality we can set a<=b<=c and thus we can set equations as such

a+b+c=Pi

0<a<Pi/2

a<=b<Pi/2

thus we can restrict this problem to what a and b are.

now there are two possible restrictions on a and b

a is in the interval (0,Pi/4] and b is in the interval [(Pi/2)-a,Pi/2)

or

a is in the interval (Pi/4,Pi/2) and b is in the interval (a,Pi/2)

now lets take a look at the minimum for the sum of the sines of a,b,c.  In either case the sum is minimized when a,b,c are minimized because sine function increases from 0 to 1 on interval [0,Pi/2].

thus in case one we can make a arbitrarily small thus we have minimum when a=0 b=Pi/2 c=Pi/2 and the sum would be 2.  but this minimum can not be reached because we can not have two right angles in a triangle.  Thus the sum is always greater than 2 in the first case

in the second case the minimum is reached when a=Pi/4 b=Pi/4 c=Pi/2 and thus the sum is 1+Sqrt(2) which is greater than 2 thus the sum is always greater than 2 in the second case as well.

Since the sum is always greater than 2 in either case then we can state that Sin(a)+Sin(b)+Sin(c)>2 for all acute triangles