The discussion revolves around a trigonometry challenge involving the angles of a triangle and a specific inequality related to their sines and cosine. Participants are exploring the proof of the inequality and discussing conditions for equality.
The discussion includes corrections and refinements, but it does not appear to reach a consensus on the proof of the inequality itself.
There are indications of missing assumptions or details in the proofs presented, and the discussion does not clarify all mathematical steps involved in establishing the inequality.
It should be :There is equality iff $C=\dfrac{2\pi}{3},\,A=B=\dfrac{\pi}{6}$.anemone said:Solution of other:
Since $\dfrac{1}{\sin x}$ is convex for $0\le x\le \pi$, we have
$\dfrac{1}{\sin A}+\dfrac{1}{\sin B}\ge \dfrac{2}{\sin \dfrac{A+B}{2}}=\dfrac{2}{\cos \dfrac{C}{2}}$
The problem will be done once we establish
$\dfrac{2}{\cos \dfrac{C}{2}}\ge \dfrac{8}{3+2\cos C}$$\dfrac{2}{\cos \dfrac{C}{2}}\ge \dfrac{8}{3+2\left(2\left(\cos \dfrac{C}{2}\right)^2-1\right)}$
$\dfrac{2}{\cos \dfrac{C}{2}}\ge \dfrac{8}{4\left(\cos \dfrac{C}{2}\right)^2+1}$
$2\left(4\left(\cos \dfrac{C}{2}\right)^2+1\right)\ge 8\cos \dfrac{C}{2}$
$\left(2\cos \dfrac{C}{2}-1\right)^2\ge 0$
There is equality iff $C=\dfrac{2\pi}{3},\,A=B=\dfrac{\pi}{3}$.