$\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). $ 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. 
 A: $AB=r\left(\cot\dfrac{A}{2}+\cot\dfrac{B}{2}\right)=r\dfrac{\sin \left(\dfrac{A+B}{2}\right)}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}}=r\dfrac{\cos \dfrac{C}{2}}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}}=r\dfrac{\sin \dfrac{π+C}{2}}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}},$ 
$A'B'=r\sqrt{2\left(1-\cos \left(\dfrac{π+C}{2}\right)\right)}=2r\sin \left(\dfrac{π+C}{4}\right)$
$\dfrac{AB}{A'B'}=\dfrac{\cos \left(\dfrac{π+C}{4}\right)}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}}$ then the inequality become:
$\dfrac{\cos \left(\dfrac{π+C}{4}\right)}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}}+\dfrac{\cos \left(\dfrac{π+B}{4}\right)}{\sin \dfrac{A}{2}\sin \dfrac{C}{2}}+\dfrac{\cos \left(\dfrac{π+A}{4}\right)}{\sin \dfrac{C}{2}\sin \dfrac{B}{2}} \ge 4\left(\sin {\dfrac{A}{2}}+\sin {\dfrac{B}{2}}+\sin {\dfrac{C}{2}}\right) $ $\iff$
$\cos \left(\dfrac{π+C}{4}\right)\sin \dfrac{C}{2}+\cos \left(\dfrac{π+B}{4}\right)\sin \dfrac{B}{2}+\cos \left(\dfrac{π+A}{4}\right)\sin \dfrac{A}{2} \ge 4\sin {\dfrac{A}{2}}\sin {\dfrac{B}{2}}\sin {\dfrac{C}{2}}\left(\sin {\dfrac{A}{2}}+\sin {\dfrac{B}{2}}+\sin {\dfrac{C}{2}}\right)$
$x=\dfrac{A}{2},y=\dfrac{B}{2},z=\dfrac{C}{2} →$
$f=\cos {\dfrac{π+2z}{4}}\sin {z}+\cos {\dfrac{π+2y}{4}}\sin {y}+\cos {\dfrac{π+2x}{4}}\sin {x} - 4\sin {x}\sin {y}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)$
$g=x+y+z-\dfrac{π}{2}$ , $F=f-\lambda g$
$F_{x}=\cos {\dfrac{π+2x}{4}}\cos {x}-\dfrac{1}{2}\sin {\dfrac{π+2x}{4}}\sin {x}-4\cos {x}\sin {y}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)-4\sin {x}\sin {y}\sin {z}\cos {x})=\lambda$ <1>
$F_{y}=\cos {\dfrac{π+2y}{4}}\cos {y}-\dfrac{1}{2}\sin {\dfrac{π+2y}{4}}\sin {y}-4\cos {y}\sin {x}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)-4\sin {x}\sin {y}\sin {z}\cos {y})=\lambda$ <2>
$F_{z}=\cos {\dfrac{π+2z}{4}}\cos {z}-\dfrac{1}{2}\sin {\dfrac{π+2z}{4}}\sin {z}-4\cos {z}\sin {y}\sin {x}\left(\sin {x}+\sin {y}+\sin {z}\right)-4\sin {x}\sin {y}\sin {z}\cos {z})=\lambda$ <3>
<1>- <2>  :
$ \sin {\dfrac{x-y}{4}}\left(\sin {\dfrac{π+2z}{8}}-3\sin {\dfrac{3z}{4}} \left(3-4\left(\sin {\dfrac{x-y}{4}}\right)^2 \right)+32\cos {\dfrac{x-y}{4}}\cos {\dfrac{x-y}{2}}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)-16\sin {x}\sin {y}\sin {z}\cos {\dfrac{x-y}{4}}\sin {\dfrac{x+y}{2}} \right)=0 $ $→ \sin {\dfrac{x-y}{4}}P_{1}\left(x,y,z\right)=0$ <4>
<2>-<3>: $\sin {\dfrac{y-z}{4}}P_{2}\left(x,y,z\right)=0$, $P_{2}\left(x,y,z\right)$ is similar as above.
$|x-y|<\dfrac{π}{2} → \cos {\dfrac{x-y}{4}}\cos {\dfrac{x-y}{2}} > 0.65 $ and $ \sin x+\sin y+\sin z \ge \sin \left(x+y+z\right)=1 → 16\cos {\dfrac{x-y}{4}}\cos {\dfrac{x-y}{2}}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)> 9\sin {\dfrac{3z}{4}}$
it is trivial that:$16\cos {\dfrac{x-y}{4}}\cos {\dfrac{x-y}{2}}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)>16\sin {x}\sin {y}\sin {z}\cos {\dfrac{x-y}{4}}\sin {\dfrac{x+y}{2}}$
$→ P_{1}\left(x,y,z\right)>0$ \sin ce it is symmetric, we have $P_{2}\left(x,y,z\right)>0$ also.
so the only solution is $x=y=z=\dfrac{π}{6}$ ,we put in $ f, f=0$,put $x=y=\dfrac{π}{2},z=0, → f>0$ so $f_{min}=0$ QED.
A: $\dfrac{\cos \left(\dfrac{π+C}{4}\right)}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}} \\ =\dfrac{2\cos \left(\dfrac{π+C}{4}\right)}{\cos \left(\dfrac{A-B}{2}\right)-\cos \left(\dfrac{A+B}{2}\right)} \\ \ge \dfrac{2\cos \left(\dfrac{π+C}{4}\right)}{1-2\sin \left(\dfrac{C}{2}\right)} \\ =\dfrac{\cos \left(\dfrac{π+C}{4}\right)}{\left(\dfrac{1}{\sqrt{2}}\cos \left(\dfrac{C}{4}\right)-\dfrac{1}{\sqrt{2}}\sin \left(\dfrac{C}{4}\right)\right)^2} \\ =\dfrac{\cos \left(\dfrac{π+C}{4}\right)}{\left(\cos \left(\dfrac{π+C}{4}\right)\right)^2}=\dfrac{1}{\cos \left(\dfrac{π+C}{4}\right)}$
$f\left(x\right)=\dfrac{1}{\cos \left(\dfrac{π+x}{4}\right)}-4\sin  \dfrac{x}{2},f''\left(x\right)=\sin \dfrac{x}{2}+\dfrac{1}{16\cos \dfrac{π+x}{4}}+\dfrac{\sin  ^2 \dfrac{π+x}{4}}{8\cos  ^3 \dfrac{π+x}{4}} > 0 \implies $
$f\left(x\right) $ is convex function $\implies \\ ∑_{\rm cyc} \left(\dfrac{1}{\cos \left(\dfrac{π+A}{4}\right)}-4\sin  \dfrac{A}{2} \right)\ge 3 f\left(\dfrac{A+B+C}{3}\right)=3f\left(\dfrac{π}{3}\right)=0$
QED.
