SOLUTION: For any triangle ABC, we have cosA + cosB + cosC =1+4SinA/2SinB/2SinC/2 and SinA/2 + SinB/2 + SinC/2=< 1/8 Use the above facts to prove that for any triangle 1,CosA+CosB+CosC=<3/

Algebra ->  Trigonometry-basics -> SOLUTION: For any triangle ABC, we have cosA + cosB + cosC =1+4SinA/2SinB/2SinC/2 and SinA/2 + SinB/2 + SinC/2=< 1/8 Use the above facts to prove that for any triangle 1,CosA+CosB+CosC=<3/      Log On


   

Question 1046557: For any triangle ABC, we have cosA + cosB + cosC =1+4SinA/2SinB/2SinC/2
and SinA/2 + SinB/2 + SinC/2=< 1/8
Use the above facts to prove that for any triangle 1,CosA+CosB+CosC=<3/2

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
I will prove an even better upper bound!
cosA+%2B+cosB+%2B+cosC+=1%2B4Sin%28A%2F2%29Sin%28B%2F2%29Sin%28C%2F2%29
===> ,
by directly substituting the given inequality...
Now A%2BB%2BC+=+180%5Eo ===> A%2F2%2BB%2F2%2BC%2F2+=+90%5Eo, which means A/2, B/2, and C/2 are acute angles, hence all sines and cosines for these angles are positive.
Now let x+=+sin%28A%2F2%29, and y+=+sin%28B%2F2%29.
===> .
The restrictions are 0 < x,y < 1.
===> z%5Bx%5D+=+4y%281%2F8-x-y%29-4xy+=+0 ===> 1%2F8-x-y-x+=+0, and y%3C%3E0.
===> 2x%2By+=+1%2F8.
Similarly, after taking z%5By%5D and solving for z%5By%5D=0 with x%3C%3E0, we get
x%2B2y+=+1%2F8.
===> the critical point is at (1/24, 1/24).
Since there is only one critical point and the domain of definition is the open square (0,1)x(0,1), there is an absolute extremum (maximum)
at the point (1/24, 1/24).
The maximum value of is thus ,
which is around 1.000289.
===> CosA%2BCosB%2BCosC+%3C=+3457%2F3456.