In an acute angled triangle, prove that $\sin A + \sin B + \sin C > 2$, where $0Is there any way to prove this? When I check intuitively by taking $A=0$ and $B=C=\frac{\pi}{2}$, the value of the expression becomes $2$ and as I changed the angles the value kept increasing in $0$ to $\frac{\pi}{2}$. I tried using Jensen's inequality but we get $\sin A + \sin B + \sin C <\frac{3\sqrt{3}}{2}$, which is of no use.
 A: $$\sin A + \sin B + \sin C=\sin A +\sin B + \sin(\pi-A-B)$$
$$=2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})+\sin(A+B)$$
$$=2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})+2\sin(\frac{A+B}{2})\cos(\frac{A+B}{2})$$
$$=2\sin(\frac{A+B}{2})\Bigl(\cos(\frac{A-B}{2})+\cos(\frac{A+B}{2})\Bigr)$$
$$=4\sin(\frac{A+B}{2})\Bigl(\cos\frac{A}{2}\cos\frac{B}{2}\Bigr)$$
$$=4\sin(\frac{\pi-C}{2})\Bigl(\cos\frac{A}{2}\cos\frac{B}{2}\Bigr)$$
$$=4\cos\frac{C}{2}\cos\frac{A}{2}\cos\frac{B}{2}$$
$$=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$$
$$0<A<B<\frac{\pi}{2}$$
$$0<\frac{A}{2}<\frac{B}{2}<\frac{\pi}{4}$$
$$4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} >4\cos\frac{C}{2}$$
when $A=0, B=0$.
During this situation, $C=\frac{\pi}{2}$
So,
$$\sin A + \sin B + \sin C=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} >4\cos\frac{\frac{\pi}{2}}{2}=2\sqrt{2}>2$$
A: I try to use comment but failed. so I just comments here. For a acute triangle, we must have all angles less than $\frac{\pi}{2}$, so $4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} $ can be direct to get the result.
