Let $A, B, C$ be angles in a triangle. Is the following inequality $4\cos A \le 1 + \cos\left(\frac{B-C}{2}\right)$ true? I just assume it but don't have a proof. Thank you for your help.
Trigonometric inequality for angles in triangle
2
$\begingroup$
geometry
trigonometry
inequality
triangles
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0Thank you N.S. This was very stupid attempt to prove the true inequality $4(\cos A + \cos B + \cos C) \le 3 + \cos(\frac{B-C}{2}) + \cos(\frac{C-A}{2}) + \cos(\frac{A-B}{2}).$ Any idea how this can be done? – 2012-07-23
1 Answers
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Nope it is not true.
The right side is at most 2, while when $A$ is very small the left side is close to 4.
Note that if $0 then
$4 \cos(A) > 2 \geq 1 + \cos\left(\frac{B-C}{2}\right)$
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0Thank you. This was very stupid attempt to prove the true inequality $4(\cos A + \cos B + \cos C) \le 3 + \cos(\frac{B-C}{2}) + \cos(\frac{C-A}{2}) + \cos(\frac{C-A}{2}).$ Any idea how this can be done? – 2012-07-23