Prove: sin A + sin B + sin C >2 if A+B+C=180 A

We'll re-write the
constraint:

A+B = 180-C

We'll
apply sine function both sides:

sin(A+B) = sin(180-C) = sin
C

We'll divide by 2 the
constraint:

(A+B)/2 =
(180-C)/2

We'll apply sine function both
sides:

sin (A+B)/2 = sin (180-C)/2 = sin (90 -C/2) = cos
C/2

Now, we'll manage the left side of the inequality,
replacing sin C by sin(A+B):

sin A + sin B + sin(A+B)
> 2

We'll transform into a product the sum sin A +
sin B:

2 sin [(A+B)/2]* cos [(A-B)/2] + sin(A+B) >
2

We'll use the double angle identity for
sin(A+B):

sin(A+B) = 2 sin [(A+B)/2]* cos
[(A+B)/2]

2 sin [(A+B)/2]* cos [(A-B)/2] + 2 sin [(A+B)/2]*
cos [(A+B)/2]>2

2 sin [(A+B)/2]* {cos [(A-B)/2] +cos
[(A+B)/2]} > 2

sin [(A+B)/2]* {cos [(A-B)/2] +cos
[(A+B)/2]} > 1

We'll transform into a product the
sum cos [(A-B)/2] +cos [(A+B)/2]:

2*sin [(A+B)/2]*cos
[(A-B+A+B)/4]*cos[(A-B-A-B)/4] > 1

2*sin
[(A+B)/2]*cos [(2A)/4]*cos[(-2B)/4] > 1

We'll
replace sin [(A+B)/2] by cos C/2, we'll simplify inside brackets and we'll cancel the
minus sign, since the cosine function is even.

cos (A/2)
*cos (B/2) *cos (C/2)> 1/2

Since the values of the
angles are smaller than 90, then all the angles are located in the 1st quadrant, where
the values of the function cosine are positive.(the direction of the inequality remains
unchanged).

We'll consider the values of the angle A = 60,
B=45 => C = 75, for the sum of the angles to be of 180
degrees.

cos 60/2* cos 45/2* cos 75/2 >
1/2

We'll use the half angle
identity:

cos 45/2 = sqrt [(1+cos
45)/2]

cos75/2 = sqrt [(1+cos
75)/2]

cos 60/2*sqrt [(1+cos 45)/2]*sqrt [(1+cos 75)/2]
> 1/2

We'll raise to square to eliminate the square
root:

3/4*[(1 + sqrt2/2)/2]*{[1+sqrt2/2*(sqrt3/2 - 1/2)]/2}
> 1/4

Computing the left side, we'll
get:

(12 + 3sqrt6 + 6sqrt2 - 3sqrt3 - 3sqrt3 - 3)/4
> 4

25.74/4 >
4

6.435 >
4

Therefore, the identity sin A + sin B + sin
C >2 is verified, under the given
constraints.