March 10th, 2008, 10:18 AM  #1 
Newbie Joined: Feb 2008 Posts: 6 Thanks: 0  Inequality with cosine
Prove that 1/(cos(A)+1/cos(B)+1/cos(C)>=6 if 0<A,B,C<90 (Angles in an acute triangle). A,B,C are of course in degrees.

June 6th, 2008, 03:44 PM  #2 
Newbie Joined: Jun 2008 From: New Zealand Posts: 20 Thanks: 0 
We need to use the following inequalities: (1) cosA + cosB + cosC <= 3/2 for more tetails about this inequality, visit http://mathcircle.berkeley.edu/archi...espdf/trig.pdf (2) (x + y + z)/3 >= cube root of xyz (AMGM Inequality) So cube root of cosAcosBcosC <= (cosA + cosB + cosC)/3 <= (3/2)/3 = 1/2 Now 1/cosA + 1/cosB + 1/cosC <= 3*cube root of [1/(cosAcosBcosC)] = 3/ cube root of (cosAcosBcosC) >= 3/(1/2) = 6 Thus 1/cosA + 1/cosB + 1/cosC >= 6 
June 7th, 2008, 05:44 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,633 Thanks: 2080 
Alternative way to start: 0 ? sin²(B  C) + (cos(B  C)  2cos A)² 0 ? sin²(B  C) + cos²(B  C)  4cos A(cos(B  C)  cos A) but cos(B  C)  cos A = cos(B  C) + cos(B + C) = 2cos(B)cos(C), so 0 ? 1  8cos(A)cos(B)cos(C), and so the cube root of cos(A)cos(B)cos(C) ? 1/2 if A, B and C are acute angles. 

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