Question

In a plane triangle ABC find maximum value of cosAcosBcosC

Answer 1

Answer:

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Step-by-step explanation:

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