IMO Hk TST

Let $D$ be the intersection point of the internal angle bisector of $\angle BAC$ and $BC$.
By the angle bisector theorem, we have $\frac{BD}{CD} = \frac{c}{b}$, hence
$$BD = \frac{ac}{b+c} \quad \text{and} \quad CD = \frac{ab}{b+c}.$$
By Stewart's theorem, we have
$$i_a^2 = \frac{1}{a} \left( b^2 \cdot \frac{ac}{b+c} + c^2 \cdot \frac{ab}{b+c} \right) - \frac{ac}{b+c} \cdot \frac{ab}{b+c} = \frac{bc((b+c)^2 - a^2)}{(b+c)^2} = \frac{4bcs(s-a)}{(b+c)^2},$$
where $s$ is the semiperimeter of $\triangle ABC$. Now, by the AM-GM inequality, since $s \neq s-a$, we have
$$i_a = \frac{2\sqrt{bc}}{b+c} \cdot \sqrt{s(s-a)} < \frac{2\sqrt{bc}}{b+c} \cdot \frac{s+(s-a)}{2} = \sqrt{bc} \le \frac{b+c}{2}.$$
Similarly, we have $i_b < \frac{c+a}{2}$ and $i_c < \frac{a+b}{2}$. Therefore, we have
$$ai_a + bi_b + ci_c < ab + bc + ca \le a^2 + b^2 + c^2.$$
This is bounded above by $9R^2$ since $9R^2 - (a^2 + b^2 + c^2) = OH^2$, where $O$ and $H$ are the circumcentre and orthocentre of $\triangle ABC$ respectively.