Can you prove the trigonometry challenge with angles of a triangle?

[anemone]

Prove that if $A,\,B$ and $C$ are angles of a triangle, then $\dfrac{1}{\sin A}+\dfrac{1}{\sin B}\ge \dfrac{8}{3+2\cos C}$.

 

[anemone]

Solution of other:

Spoiler

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 $C=\dfrac{2\pi}{3},\,A=B=\dfrac{\pi}{6}$.

 

[Albert1]

Solution of other:

Spoiler

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 $C=\dfrac{2\pi}{3},\,A=B=\dfrac{\pi}{3}$.

It should be :There is equality iff $C=\dfrac{2\pi}{3},\,A=B=\dfrac{\pi}{6}$.

 

[anemone]

Thanks, Albert for catching it...

I've hence fixed the mistake.