$\sin{\frac{A+B}{2}}+\sin{\frac{B+C}{2}}+\sin{\frac{C+A}{2}} > \sin{A}+\sin{B}+\sin{C}. $ Help me please to prove that: 
for any $\triangle ABC$ we have the following inequality: $$\sin{\frac{A+B}{2}}+\sin{\frac{B+C}{2}}+\sin{\frac{C+A}{2}} > \sin{A}+\sin{B}+\sin{C}. $$
It's about convexity ? 
thanks :)

Hint: Eliminate $\angle C$
$A+B+C=180^0 \implies C=180-(C+B)$
Now you just need to show :
$\sin{\frac{A+B}{2}}+\sin{(90-\dfrac{A}{2})}+\sin{(90-\dfrac{B}{2})} \ge \sin{A}+\sin{B}+\sin{(180-(A+B))}$
$\sin{\frac{A+B}{2}} +\cos{\frac{B}{2}}+\cos{\frac{A}{2}} \ge \sin A+ \sin B+ \sin(A+B)$
