Trigonometric inequality for angles in triangle

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.

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Thank 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? – Martin Jul 23 '12 at 15:34

The right side is at most 2, while when $A$ is very small the left side is close to 4.
Note that if $0 <A< 30^\circ$ then
$$4 \cos(A) > 2 \geq 1 + \cos\left(\frac{B-C}{2}\right)$$
Thank 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? – Martin Jul 23 '12 at 15:24