Indija TS 2016

Rotate the triangle around the centre of $\Gamma$ by $180^{\circ }$. Then $A_1$, $B_1$, $C_1$ respectively go to $A_2$, $B_2$, $C_2$. These are respectively the midpoints of the minor arc $BC$, $CA$ and $AB$. We see that $A_1{B}_1{C}_1$ and $A_2{B}_2{C}_2$ are congruent triangles and hence have the same in-radii. But we know that $A_2$ is the intersection of the angle bisector of $\angle BAC$ with $\Gamma$ and similar descriptions hold for the other two points. ---

Observe that $\angle C_2{A}_{2}A=\angle C_{2}CA=\angle C/2$ and $\angle B_2{A}_{2}A=\angle B_{2}BA=\angle B/2$. Hence $\angle A_2=(\angle B/2)+(\angle C/2)=(\pi -\angle A)/2$. Similarly we get other angles: $$\angle B_2=(\pi -\angle B)/2,\angle C_2=(\pi -\angle C)/2.$$ We know that for any triangle $ABC$ its inradius is given by $$r=4R\sin (A/2)\sin (B/2)\sin (C/2),$$ where $R$ is its circumradius. If $r$ and $r_1$ are respectively the inradii of $△ABC$ and $△A_2{B}_2{C}_2$, then we have $$\frac{r}{r_1}=\frac{\sin (A/2)\sin (B/2)\sin (C/2)}{\sin ((\pi -A)/4)\sin ((\pi -B)/4)\sin ((\pi -C)/4)}.$$ Now we use another known identity in a triangle: $$1+4\prod \sin \left(\frac{\pi -A}{4}\right)=\sum \sin \frac{A}{2}.$$ Thus we obtain $$\frac{r}{r_1}=\frac{4\prod \sin (A/2)}{(\sum \sin (A/2))-1}.$$ We have to show that $r\le r_1$. Equivalently we have to prove that $$1+4\prod \sin (A/2)\le \sum \sin (A/2).$$ But we know that $$\sum \sin (A/2)\ge 3{\left(\prod \sin (A/2)\right)}^{1/3}.$$ Thus it is sufficient to prove that $$1+4\prod \sin (A/2)\le 3{\left(\prod \sin (A/2)\right)}^{1/3}.$$ Introducing $x={\left(\prod \sin (A/2)\right)}^{1/3}$, this inequality is equivalent to $$4x^3-3x+1\ge 0.$$ However it is easy to see that $4x^3-3x+1=(2x-1{)}^2(x+1)$. Hence $4x^3-3x+1\ge 0$ for all positive $x$. This completes the proof.