sin(alpha)+sin(beta)+sin(gamma)-sin(alpha+beta+gamma)
=2sin((alpha+beta)/2)cos((alpha-beta)/2)-(sin(alpha+beta+gamma)-sin(gamma))
=2sin((alpha+beta)/2)cos((alpha-beta)/2)-(2cos(alpha/2+beta/2+gamma)sin((alpha+beta)/2))
=2sin((alpha+beta)/2)[cos((alpha-beta)/2)-cos(alpha/2+beta/2+gamma)]
=2sin((alpha+beta)/2)[2sin((alpha+gamma)/2)sin((beta+gamma)/2)]>0
As it is given
0 < alpha,beta and gamma< pi/2 we have,
((alpha+beta)/2),((alpha+gamma)/2)and((beta+gamma)/2) arr acute angles and sine of all must be > 0
Hence
sin(alpha)+sin(beta)+sin(gamma) >sin(alpha+beta+gamma)