Note: Using the cosine rule, this inequality can be written as:
Now, we have to prove that $cosA+cosB+cosC\le3/2$
$f(A,B) = cos(A) + cos(B) + cos(π-A-B)$
$f(A,B) = cos(A) + cos(B) - cos(A+B)$
within the region R:
Since it's an open region, we know the maximum cannot occur anywhere along the boundary of R and must occur at some critical point(s) in the interior.
So we just have to find the point where the total derivative is 0. Of course, A and B aren't dependent on each other, so we can just set the two partials equal to zero to find the critical point(s):
$$sin(A+B) - sin(A) = 0$$
$$sin(A+B) - sinB = 0$$
So $sinA = sinB$, thus A=B
But $sinA = sin2A = 2sinAcosA$
Thus $$cosA = 1/2$$
Note: we can divide by sinA because A≠0. This also excludes A=0 as a solution, since A=0 is in the boundary of R, which is not a part of our valid region.
$$A = π/3$$
So the maximum is at $$A=B=C=π/3$$.
Plugging that in, we know the maximum value of f is:
$3cos(π/3) = 3/2$
Thus for angles A,B,C of a triangle:
$$cos(A)+cos(B)+cos(C) = f(A,B) = ≤ 3/2$$ .