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I need help proving the following trig inequality:

$$\cos a-\cos b-\cos c\geq -\frac{3}{2}$$ where $a+b+c=2\pi$ and $a,b,c>0$.

I've found that equality occurs when $a=\frac{4\pi}{3},b=c=\frac{\pi}{3},$ but I'm not sure how to prove it in general.

I would also like to only use math covered in precalc and below, if that matters.

Blue
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HelloPeople99
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  • Maybe cosine rule will simplify it a bit. – Ralph Clausen Aug 13 '20 at 17:55
  • https://math.stackexchange.com/questions/1382047/trigonometry-problem/3267312#3267312 or https://math.stackexchange.com/questions/1419652/prove-that-in-any-triangle-abc-cos2a-cos2b-cos2c-geq-frac34 or https://math.stackexchange.com/questions/1374163/in-a-triangle-find-the-minimum-and-maximum-of-cosa-b-cosb-c-cosc-a?rq=1 – lab bhattacharjee Aug 13 '20 at 19:21

2 Answers2

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We need to prove that $$\cos{a}-2\cos\frac{b+c}{2}\cos\frac{b-c}{2}+\frac{3}{2}\geq0$$ or $$2\cos^2\frac{a}{2}-1+2\cos\frac{a}{2}\cos\frac{b-c}{2}+\frac{3}{2}\geq0$$ or $$4\cos^2\frac{a}{2}+4\cos\frac{a}{2}\cos\frac{b-c}{2}+1\geq0$$ or $$\left(2\cos\frac{a}{2}+\cos\frac{b-c}{2}\right)^2+\sin^2\frac{b-c}{2}\geq0,$$ which is obvious.

Michael Rozenberg
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Put $t = \cos\left(\frac{b+c}{2}\right) \implies LHS = 2t^2 -2t\cos\left(\frac{b-c}{2}\right) + \dfrac{1}{2} \ge 0$. But this is clear viewed as a quadratic in $t$ and equality occurs at $\cos\left(\frac{b-c}{2}\right) = 1 \implies b = c ...$

DeepSea
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