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How to find maximum value cosAcosBcosC in the plane triangle
If any of A, B and C are not acute, then clearly the product is non-positive (since cosx ≤ 0 for π/2 ≤ x ≤ π, and at most one angle can be non-acute. Thus we have the product of 2 positive numbers and a non-positive number, which is thus non-positive) Thus if any of A, B, and C are not acute then cosA cosB cosC ≤ 0 Now if they are all acute then we have that cosA, cosB and cosC are all positive. Thus, by AM-GM, we have that (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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