Prove that in an acute angled triangle ABC, cos A cos B cos C ≤ 1/8

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Q: Prove that in an acute angled triangle ABC, cos A cos B cos C ≤ 1/8

Solution : Let 8 (cos A cos B cos C) = y

4 (2 cos A cos B ) cos C = y

4 [cos(A + B) + cos(A-B)] cosC = y

(Since ,  A+B+C = π)

4 [cos(π-C) + cos(A-B)] cosC = y

– 4 cos2C + 4 cosC . cos(A-B) = y

4 cos2C – 4 cosC . cos(A-B) + y = 0

$\displaystyle cosC = \frac{-4cos(A-B) \pm \sqrt{16 cos^2 (A-B) – 16 y}}{8}$

Since cos C is real

y ≤ cos2 (A – B)

⇒ y ≤ 1

⇒ cos A cos B cos C ≤ 1/8

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