Final Round of National Olympiad

Let $|BC| = a$, $|CA| = b$ and $|AB| = c$ and the angles opposite to those sides be $\alpha$, $\beta$ and $\gamma$, respectively (see fig. 24).

Figure 24

The law of sines in triangles $ABC$ and $IBC$ gives $\frac{a}{\sin \alpha} = 2R$ and $\frac{a}{\sin \left(\frac{\beta}{2} + \frac{\gamma}{2}\right)} = 2R_A$, respectively. As $\sin\left(\frac{\beta}{2} + \frac{\gamma}{2}\right) = \sin\left(90^\circ - \frac{\alpha}{2}\right) = \cos\frac{\alpha}{2}$, we obtain $\frac{R_A}{R} = \frac{\sin\alpha}{\cos\frac{\alpha}{2}} = 2\sin\frac{\alpha}{2}$. Similarly we get $\frac{R_B}{R} = 2\sin\frac{\beta}{2}$ and $\frac{R_C}{R} = 2\sin\frac{\gamma}{2}$.
To solve the problem we need to prove that $\frac{R_A}{R} + \frac{R_B}{R} + \frac{R_C}{R} \le 3$, or equivalently,
$$\sin \frac{\alpha}{2} + \sin \frac{\beta}{2} + \sin \frac{\gamma}{2} \le \frac{3}{2}. \qquad (6)$$
Applying Jensen's inequality gives
$$\frac{1}{3} \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} + \sin \frac{\gamma}{2} \right) \le \sin \left( \frac{\frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2}}{3} \right) = \sin \left( \frac{\alpha + \beta + \gamma}{6} \right) = \sin \frac{\pi}{6} = \frac{1}{2},$$
which directly implies the necessary result.