Question

Maximum value of cosa cosb cosc in a plane triangle

Answer 1

Hey mate here is ur query


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Thus if any of A, B, and C are not acute then 
cosA cosB cosC ≤ 0 

so we ignore upper case


Now if they are all acute then we have that cosA, cosB and cosC are all positive. 
Thus, by AM-&GM,

we have that
As AM >= GM
(cosA + cosB + cosC)/2 ≥ (cosA cosB cosC)^(1/3) 
But we know that cosA + cosB + cosC ≤ 3/2 by problem 1. 
Thus (cosA cosB cosC)^1/3 ≤ (3/2)/3 = 1/2 
Cubing each side we get 


cosA cosB cosC ≤ 1/8 
(
We can do this since f(x) = x^3 is increasing) 

Again, we have equality if the triangle is equilateral. 
Thus the maximum value of

cosA cosB cosC is 1/8 :)

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Hope this will help you