# Prove that 2*sqrt3*(sinA+sinB+sinC)/9 < 1

*print*Print*list*Cite

### 2 Answers

We know that sinA+sinB+sinC = 4cosA/2*cosB/2*cosC/2..........(1), where A .B and C are angles of a triangle.

{ cosA/2*cosB/2*cosC/2}^(1/3) <= {[cosA/2+cosB/2+cosc/3]/3}..(2) as GM < AP holds good as cosA/2 , cosB/2 and cosC/2 are all postive for A/2, B/2 and C/2 each of which are less than 180/2 =90 deg.

The equality sign holds only when A/2 = B/2 = C/2 = 30 deg when RHS of eq (2) is {sqr3/2+sqrt3/2+sqrt3/2}/3 = sqrt3/2. So,

cosA/2*cosB/2*cosC/2 < ((sqrt3)/2)^3. Substitute this in (1):

sinA+sinB+sinC < 4 ((sqrt3)/2)^3 = (3sqrt3)/2 = (9/2)/sqrt3.

Divide both sides by (9/2) /sqrt3 and get:

2sqrt3(sinA+sinB+sinC) /9 < 1.

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**