In an Acute angled Triangle $\Delta ABC$ $\csc \left(\frac{A}{2}\right)+\csc \left(\frac{B}{2}\right)+\csc \left(\frac{C}{2}\right)=6$

Question

In an Acute angled Triangle $\Delta ABC$ if $$\csc \left(\frac{A}{2}\right)+\csc \left(\frac{B}{2}\right)+\csc \left(\frac{C}{2}\right)=6$$ Prove that the Triangle is Equilateral.

I tried using $A.M \ge H.M$ So we have

$$\frac{\sin \left(\frac{A}{2}\right)+\sin\left(\frac{B}{2}\right)+\sin \left(\frac{C}{2}\right)}{3} \ge \frac{3}{\csc \left(\frac{A}{2}\right)+\csc \left(\frac{B}{2}\right)+\csc \left(\frac{C}{2}\right)}=\frac{1}{2}$$ $\implies$

$$\sin \left(\frac{A}{2}\right)+\sin\left(\frac{B}{2}\right)+\sin \left(\frac{C}{2}\right) \ge \frac{3}{2}$$

Then i started to analyze the equality case i.e.,

$$\sin \left(\frac{A}{2}\right)+\sin\left(\frac{B}{2}\right)+\sin \left(\frac{C}{2}\right)=\frac{3}{2} \tag{1}$$

Let $x=\sin \left(\frac{A}{2}\right)$ and $y=\sin \left(\frac{B}{2}\right)$

Then $$\sin \left(\frac{C}{2}\right)=\sin \left(\frac{180-A-B}{2}\right)=\cos \left(\frac{A+B}{2}\right)=\sqrt{1-x^2}\sqrt{1-y^2}-xy$$

Then we have from $(1)$

$$f(x,y)=x+y-xy+\sqrt{1-x^2}\sqrt{1-y^2}-\frac{3}{2}$$

I partially differentiated with respect to $x$ and got

$$y=1-2x^2$$ $\implies$

$$\sin \left(\frac{B}{2}\right)=\cos A$$

Can we proceed further?

Answer 1

Let $f(x)=\frac{1}{\sin\frac{x}{2}}$.

Thus, $f''(x)=\frac{3+\cos{x}}{8\sin^3\frac{x}{2}}>0$ for all $x\in\left(0,\frac{\pi}{2}\right)$ and by Jensen we obtain: $$\sum_{cyc}\frac{1}{\sin\frac{\alpha}{2}}\geq\frac{3}{\sin\frac{\alpha+\beta+\gamma}{6}}=6.$$ The equality occurs for $\alpha=\beta=\gamma=60^{\circ}$ only and we are done.