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March 10th, 2008, 10:18 AM   #1
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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.
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June 6th, 2008, 03:44 PM   #2
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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 (AM-GM 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
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June 7th, 2008, 05:44 AM   #3
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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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