, 2004

Solution:
We know that $2R = \frac{abc}{2\triangle}$ and $r_{a} = \frac{\triangle}{s-a}$, where $a, b, c$ are the sides of the triangle $ABC$, $s = \frac{a+b+c}{2}$ and $\triangle$ is the area of $ABC$. Thus the given condition $2R \leq r_{a}$ translates to
$$abc \leq \frac{2\triangle^{2}}{s-a}$$
Putting $s-a = p$, $s-b = q$, $s-c = r$, we get $a = q + r$, $b = r + p$, $c = p + q$ and the condition now is
$$p(p+q)(q+r)(r+p) \leq 2\triangle^{2}$$
But Heron's formula gives, $\triangle^{2} = s(s-a)(s-b)(s-c) = pqr(p+q+r)$. We obtain $(p+q)(q+r)(r+p) \leq 2qr(p+q+r)$. Expanding and effecting some cancellations, we get
$$p^{2}(q+r) + p(q^{2} + r^{2}) \leq qr(q+r)$$
Suppose $a \leq b$. This implies that $q + r \leq r + p$ and hence $q \leq p$. This implies that $q^{2} r \leq p^{2} r$ and $q r^{2} \leq p r^{2}$ giving $qr(q+r) \leq p^{2} r + p r^{2} < p^{2} r + p r^{2} + p^{2} q + p q^{2} = p^{2}(q+r) + p(q^{2} + r^{2})$ which contradicts $(\star)$. Similarly, $a \leq c$ is also not possible. This proves (i).

Suppose $2R \leq r_{b}$. As above this takes the form
$$q^{2}(r+p) + q(r^{2} + p^{2}) \leq pr(p+r)$$
Since $a > b$ and $a > c$, we have $q > p$, $r > p$. Thus $q^{2} r > p^{2} r$ and $q r^{2} > p r^{2}$. Hence
$$q^{2}(r+p) + q(r^{2} + p^{2}) > q^{2} r + q r^{2} > p^{2} r + p r^{2} = pr(p+r)$$
which contradicts $(\star\star)$. Hence $2R > r_{b}$. Similarly, we can prove that $2R > r_{c}$. This proves (ii).