Prove that in any triangle ABC, 8.cos(A).cos(B).cos(C) < 1.

This "geometry" problem can be treated fairly readily with advanced calculus through use of a Lagrange multiplier.  The constraint is g=A+B+C-pi=0 and we wish to maximize f=cosA*cosB*cosC. Since grad(g)=(1,1,1)~=0, we may apply the Lagrange multiplier method where grad(f)=L*grad(g), or in our case,

(-sinA*cosB*cosC,-sinB*cosA*cosC,-sinC*cosA*cosB)=(L,L,L)

determines the critical points, L being the Lagrange multiplier.  Hence cotA=cotB=cotC, and because the cotangent function is one-to-one on (0,pi), this is satisfied only for A=B=C=pi/3 where f then equals 1/8. The domain of definition is A+B+C=pi, A,B,C>0, and on it's boundary f can only take values between -1 and 0 since setting C=0, say, makes B=pi-A, cosB=-cosA, and f=-(cosA)^2. Hence the maximum value of f on its domain of definition must be the 1/8 that it attains at it's only critical point.

Edited on March 27, 2005, 1:45 am

Posted by Richard on 2005-03-23 22:55:31