Prove that

CosA + CosB + CosC 3/2 for a triangle ABC

( means less than or equal to)

Since C = π - A - B, we want to maximize the function:

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:

A+B<π
0<A,B

Since it's an open region, we know the max cannot occur anywhere along the boundary of Rand 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) - sin(B) = 0

So sin(A) = sin(B), thus A=B

But sin(A) = sin(2A) = 2sin(A)cos(A)

Thus cos(A) = 1/2

Note: we can divide by sin(A) 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 max 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 .

Let cosA+cosB+cosC=P2cos(A/2+B/2)cos(A/2-B/2)+1-2sin^c/2=P2sinC/2.cos(A/2-B/2)+1-2sin^C/2=PRearrange it properly2sin^C/2-2sinC/2.cos(A/2-B/2)-1+P=0Quadratic equation in sinC/2 So ∆>=04cos^(A/2-B/2)>=4×2×(P-1)Max value of Cos^(A/2-B/2) is 1Therefore 1>=2P-2 2P