Prove that 2*sqrt3*(sinA+sinB+sinC)/9

We'll put A,B,C as being the angles of the triangle
ABC.

We'll associate a function f(x)=sin
x

The variable x belongs to the interval [0,pi](because of
the constraint that A,B,C are the angles of the triangle ABC, where the sum of the
angles is 180 degrees).

We'll prove the
requested inequality, using Jensen's inequality.

We'll
calculate the first and the second derivative of
f(x):

f'(x)=cos x

f"(x)=-sin
x<0 => the function is concave

Because the
function is concave we'll apply the Jensen's
inequality:

f[(A+B+C)/3]>[f(A)+f(B)+f(C)]/3

Working
in a triangle, A+B+C=180 and (A+B+C)/3=180/3=60

f(60)=sin
60=(sqrt 3)/2

[f(A)+f(B)+f(C)]/3=(sin A+sin B+sin
C)/3

We'll substitute the resulted values in Jensen's
inequality:

(sin A+sin B+sin C)/3<(sqrt
3)/2

sin A+sin B+sin C<3(sqrt 3)/2

We'll cross
multiply:

2*(sin A+sin B+sin C) < 3(sqrt
3)

2*(sin A+sin B+sin C) / 3(sqrt 3) <
1

2*sqrt 3*(sin A+sin B+sin C) / 3*3 <
1

2*sqrt
3*(sin A+sin B+sin C) / 9 <
1