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Question: Cosine inequality Prove that in a triangle ABC the following inequality holds: cos A cos B cos C <= 1/8. Answer: First Proof This proof is given by Ron L.J. van den Burg. Note that A+B+C = Pi, or C = Pi - A - B. Let P(A,B) = cos A cos B cos (Pi - A - B). Note that P(A,B) is the product on the left of the inequality. Since P(A,B) is continuous, differentiable and bound for all real A and B, candidates for maxima are found where d P(A,B) / d A = 0 and d P(A,B) / d B = 0. The derivative of P with respect to A is dP/dA (A, B) = cos B ( sin A cos (Pi - A - B) - cos A sin (Pi - A - B) ). dP/dA (A, B) = 0 if B = Pi/2 or sin A : cos A = sin (Pi - A - B) : cos (Pi - A - B). This is equivalent to B = Pi/2 or A = Pi - A - B = C. Similarly, the derivative of P with respect to B is zero if A = Pi/2 or B = C. Let's check candidates for maxima A = Pi/2 or B = Pi/2. Both candidates A = Pi/2 and candidates B = Pi/2 have a value for P(A,B) of zero, while P(Pi/3, Pi/3) equals 1/8, so these candidates are not maxima. The only candidate remaining is A = B = C = Pi/3 for which P(A,B) = 1/8. Concluding: the maximum over all A and B of cos A cos B cos C where C = Pi - A - B is 1/8 so cos A cos B cos C is less than or equals to 1/8. - Ron L.J. van den Burg. Second proof One can also prove the above result without calculus by transforming the expression 1 - 8*cos A cos B cos C into a sum of squares by using trig functions. The product-sum formula gives 1 - 8*cos A cos B cos C = 1 - 4*cos A * (cos (B-C) + cos (B+C)). Now we note that cos (B+C) = -cos(pi - B - C) = -cos A. Hence The expression above is equal to (sin (B-C))^2 + (cos (B-C))^2 - 4*cos A cos (B - C) + 4*(cos A)^2 = (sin (B-C))^2 + (cos (B-C) - 2*cos A)^2 since (sin (B-C))^2 + (cos (B-C))^2 = 1. The expression is nonnegative, so 1 - 8*cos A cos B cos C >= 0 and the result follows. - Henry Shin |
