, 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 $A_1B_1C_1$ and $A_2B_2C_2$ 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.

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Observe that $\angle C_2A_2A = \angle C_2CA = \angle C/2$ and $\angle B_2A_2A = \angle B_2BA = \angle B/2$. Hence $\angle A_2 = (\angle B/2) + (\angle C/2) = (\pi - \angle A)/2$. Similarly we get other angles:
$$\angle B_2 = (\pi - \angle B)/2, \quad \angle C_2 = (\pi - \angle C)/2.$$
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 $\triangle ABC$ and $\triangle A_2B_2C_2$, 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.