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?

MaxWong Jun 22, 2019

#1

#2**0 **

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.

MaxWong
Jun 22, 2019

#3**-4 **

Try this:

If A is fixed then B+C is fixed, and the product is greatest when B=C

CuteDramione
Jun 22, 2019

#6**0 **

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?

MaxWong
Jun 22, 2019