Harvard-MIT Mathematics Tournament

Solution:
It is well-known that both $\frac{a b c}{4 R}$ and $\frac{r(a+b+c)}{2}$ are equal to the area of triangle $A B C$. Thus $\frac{a b c}{4 R}=\frac{r(a+b+c)}{2}$, and
$$R r=\frac{a b c}{2(a+b+c)}$$
Since $a \leq b$ and $a \leq c$, we have $\frac{a^{2}}{b c} \leq 1$. We thus obtain that
$$\begin{aligned} \frac{a x}{r R} & =\frac{a(b+c-a) / 2}{\frac{a b c}{2(a+b+c)}} \\ & =\frac{(a+b+c)(b+c-a)}{b c} \\ & =\frac{(b+c)^{2}-a^{2}}{b c} \\ & =\frac{(b+c)^{2}}{b c}-\frac{a^{2}}{b c} \\ & =\frac{b}{c}+\frac{c}{b}+2-\frac{a^{2}}{b c} \\ & \geq \frac{b}{c}+\frac{c}{b}+2-1 \\ & \geq 2+2-1 \\ & =3 \end{aligned}$$
Equality is achieved when $a=b=c$.