$\frac{AB}{A'B'}+\frac{BC}{B'C'}+\frac{CA}{C'A'} \geq 4 \left(\sin{\frac{A}{2}}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}}\right). $

Question

Let be a circle inscribed in the triangle $\triangle ABC$ wiht the center $I$. The intersection of the circle with $AI$ is $A'$, with $BI$ is $B'$ and with $CI$ is $C'$.

Prove that: $$\frac{AB}{A'B'}+\frac{BC}{B'C'}+\frac{CA}{C'A'} \geq 4 \left(\sin{\frac{A}{2}}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}}\right). $$

thanks. seems to be to hard for me.