Let be positive real numbers such that . Suppose that the maximum of is of the form , where are relatively prime integers. Find .
This problem was inspired by problem from Vietnamese Mathematical Olympiads 1999.
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From b c − c a − a b = 1 , we can write it as a ⋅ b 1 + b 1 ⋅ c 1 + c 1 ⋅ a = 1 . The fact is that x y + y z + z x = 1 if and only if there exist 0 < A , B , C < π such that A + B + C = π and x = tan 2 A , y = tan 2 B , z = tan 2 C . Here, we take x = a , y = b 1 , z = c 1 . So, we have that a = tan 2 A , b = cot 2 B , c = cot 2 C . Let P be the expression we want to maximize. We can alternatively write it as P = 4 0 2 4 cos 2 2 A − 4 0 2 4 sin 2 2 B − 2 5 5 5 sin 2 2 C = 2 0 1 2 ( 1 + cos A ) − 2 0 1 2 ( 1 − cos B ) − 2 5 5 5 sin 2 2 C = 2 0 1 2 ( cos A + cos B ) − 2 5 5 5 sin 2 2 C = 4 0 2 4 cos 2 A + B cos 2 A − B − 2 5 5 5 sin 2 2 C = 4 0 2 4 cos 2 π − C cos 2 A − B − 2 5 5 5 sin 2 2 C = 4 0 2 4 sin 2 C cos 2 A − B − 2 5 5 5 sin 2 2 C . Let u = sin 2 C and v = cos 2 A − B . We then have that P = 4 0 2 4 u v − 2 5 5 5 u 2 = − 2 5 5 5 ( u − 2 5 5 5 2 0 1 2 v ) 2 + 2 5 5 5 2 0 1 2 2 v 2 , whose maximum is P max = 2 0 1 5 2 0 1 2 2 , which occurs when u − 2 0 1 5 2 0 1 2 v = 0 , and v = 1 . Working on variables, we'll finally find out that the maximum occurs when ( a , b , c ) = ( a 0 , b 0 , c 0 ) , where a 0 = 4 5 6 7 5 4 3 , b 0 = 5 4 3 4 5 6 7 , c 0 = 2 0 1 2 5 4 3 ⋅ 4 5 6 7 . This is to confirm that such maximum exists. Now, our task is to find the P max , which is 2 5 5 5 2 0 1 2 2 . One can check that g cd ( 2 0 1 2 , 2 5 5 5 ) = 1 , so that the answer is 2 0 1 2 + 2 5 5 5 = 4 5 6 7 .