Let a, b and c be the angles in a triangle. Show that:

sin(a) + sin(b) + sin(c) <= 3√3/2

Suppose that 3√3/2 is not the maximum value over all possible triplets of angles a, b and c with a + b + c = π. Let M be this maximum value, and choose a, b and c with a + b + c = π and sin(a) + sin(b) + sin(c) = M > 3√3/2.

Since sin(π/3) + sin(π/3) + sin(π/3) = 3√3/2, not all of a, b and c can be equal to π/3. Furthermore, one of a, b or c must be greater than π/3, since if they were all less than or equal to π/3, and one was not equal to π/3, then the sum would be less than π. Similarly, one of a, b or c must be less than π/3.

Relabel the angles so that a is the smallest and c is the largest. Then we have a < π/3 and c > π/3.

Now let r = (c-a)/2 and x = (c+a)/2. r is greater than zero, c = x + r and a = x - r. Also, x + b + x = a + r + b + c - r = a + b + c = π, so x, x and b make up three angles of a triangle. Furthermore:

sin(a) + sin(b) + sin(c)

= sin(x-r) + sin(b) + sin(x+r)

= sin(x)cos(r) - cos(x)sin(r) + sin(b) + sin(x)cos(r) + cos(x)sin(r)

= 2sin(x)cos(r) + sin(b)

< sin(x) + sin(x) + sin(b)

since r > 0 and hence cos(r) < 1.

Now we have found three angles x, x and b whose sum is π, and for which sin(x) + sin(x) + sin(b) > sin(a) + sin(b) + sin(c). Hence sin(a) + sin(b) + sin(c) is not maximal!

This contradiction establishes that it is impossible to find a, b and c with a + b + c = π, and sin(a) + sin(b) + sin(c) > 3√3/2. And so the maximum value of sin(a) + sin(b) + sin(c) with a + b + c = π must equal 3√3/2, since this value is attained when a = b = c = π/3.