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/
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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) (Show Source): You can put this solution on YOUR website!
I will prove an even better upper bound!
===> ,
by directly substituting the given inequality...
Now ===> , which means A/2, B/2, and C/2 are acute angles, hence all sines and cosines for these angles are positive.
Now let , and .
===> .
The restrictions are 0 < x,y < 1.
===> ===> , and .
===> .
Similarly, after taking and solving for with , we get
.
===> 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.
===> .
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