prove that (cosA*cosB*cosC)<=1/8

Asked by  | 14th Dec, 2009, 12:11: PM

Expert Answer:

A, B,C are the angles of the triangle then A+B+C= 180o

If we consider the triangle to be acute angled and each angle to be 60o   then  cos A + cos B+CosC=3/2

 In general cos A + cos B+CosC3/2
use A.M G.M
(cos A +CosB+cosC)/3 (cosA.cosB.cosC)^1/3
so, cosA.cosB.cosC 1/8

Answered by  | 16th Dec, 2009, 09:14: AM

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