In \(\triangle ABC\), what is the maximum value of \(\cos A + \cos B + \cos C\)?
I have solved the counterpart of this question with sin instead of cos. I used Jensen's inequality to get the result that the maximum occurs at A = B = C = pi/3. However, as the concavity of y = cos x changes at x = pi/2, Jensen's inequality does not work anymore. Can someone help?
Nope. I am sure it could be greater.
For example. in a 90-45-45 triangle, cos A + cos B + cos C = 2 sqrt (2), which is already greater than sqrt(3)/2.
If A is fixed then B+C is fixed, and the product is greatest when B=C
If A is fixed, max value occurs when B = C,
If B is fixed, max value occurs when A = C.
If C is fixed, max value occurs when A = B.
So the maximum will occur when A = B = C
Max value = 3 cos (pi / 3) = 3/2
I got the answer, but why does the max value occur when B = C if A is fixed?