If 0<(alpha),(beta),(gamma)<(pi)/2 prove that sin(alpha)+sin(beta)+sin(gamma)>sin(alpha+beta+gamma)?

1 Answer
Aug 5, 2018

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)