Inequality in triangle: $2\cos(A)\le\cos(B-C)$

Question

Let $ABC$ be an acute-angled triangle. If $A \geq B \geq C$, show that $$2 \cos(A) \leq \cos(B-C).$$

I've tried all sorts of trigonometric relationships to get this to a nicer form, but I couldn't arrive at anything manageable. I have observed that equality is acquired when $ABC$ is equilateral. Any tips?

Answer 1

Given $\pi > A\geq B\geq C > 0$ and $A+B+C=\pi$, we can know

1. $\cos(A) \leq \cos(B)$
2. $\cos(C) \geq \cos(2\pi/3) = 0.5$

So $$ \cos(A) \leq \cos(B) \leq 2\cos(B)\cos(C) = \cos(B-C) + \cos(B+C) = \cos(B-C) - \cos(A) $$

Thus $$ 2\cos(A) \leq \cos(B-C) $$

Answer 2

Hint: Using the Theorem of cosines and the additionformula we get $$2\frac{b^2+c^2-a^2}{2bc}\le \frac{(a^2+c^2-b^2)(b^2+a^2-c^2)}{4a^2bc}+\frac{4A^2}{a^2bc}$$ Now using that $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ and we obtain after simplifications: $$1/2\,{\frac {2\,{a}^{4}-{a}^{2}{b}^{2}-{a}^{2}{c}^{2}-{b}^{4}+2\,{b}^{ 2}{c}^{2}-{c}^{4}}{{a}^{2}bc}} \geq 0$$

Answer 3

Let $B=\frac{\pi}{3}+x$, $C=\frac{\pi}{3}-y$, so $A=\frac{\pi}{3}+y-x$. Then the condition $A\ge B$ implies $y\ge2x$, that is $y-x\ge x$. Also $0\le y\le\pi/3$, so $$\sin x\sin y\le\frac{\sqrt3}{2}\sin(y-x)$$ Hence $$\cos(x-y)-\cos(x+y)\le\sqrt3\sin(y-x)$$ $$\therefore \cos(x-y)-\sqrt3\sin(y-x)\le\cos(x+y)$$ But this is just $$2\cos A\le\cos(B-C)$$