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Is the following inequality in a triangle known? $$4(\cos A + \cos B + \cos C) \le 3 + \cos \left(\frac{B-C}{2}\right) + \cos \left(\frac{C-A}{2}\right) + \cos \left(\frac{A-B}{2}\right)$$ It looks correct to me but I would appreciate if someone confirm it.

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    It's not correct. Let $a=0, b=\pi/2, c=1$2012-07-24
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    The assumption is that $A+B+C = \pi$, i.e. $A,B,C$ are angles in a triangle.2012-07-24
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    Evaluating the inequality with Mathematica with brute force, for $0, $B\leq A<\pi-B$ and $C=\pi-A-B$ and with a grid of tickness $\pi/1000$, it seems to be true, but I have no proof. Moreover, every relaxation on the coefficients (4.1 instead of 4, or 2.9 instead of 3) invalidates the inequality.2012-07-24
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    Thank you enzotib. Your mathematical experiment gives strong enough evidence. It appears to be difficult to prove the inequality rigorously.2012-07-24
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    The inequality is strict at $A=B=C=\pi/3$ and permutations of $A=B=0, C=\pi$, so relaxation of the coefficients is definitely not possible.2012-07-25
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    To aid the intuition, I made some ternary plots of the quantities in question with respect to $A$, $B$, and $C$. Here are [the left-hand side ($l$)](http://i.stack.imgur.com/6iLCg.png) and [the right-hand side ($r$)](http://i.stack.imgur.com/lSqUF.png), both of which appear to lie between $4$ and $6$. [Their difference $r-l$](http://i.stack.imgur.com/tIOTj.png) is pretty ugly, but maybe [the ratio $(l-4)/(r-4)$](http://i.stack.imgur.com/q4Ab3.png) can be shown to be concave.2012-07-26
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    This is equivalent to http://math.stackexchange.com/questions/783189/a-geometric-inequality-proving-8r2r-le-am-1bm-2cm-3-le-6r and you have two proofs there :)2014-05-22

4 Answers 4