Given is a triangle $ABC$, the inscribed circle $G$ of which has radius $r$. Let $r_a$ be the radius of the circle touching $AB$, $AC$ and $G$. [This circle lies inside triangle $ABC$.] Define $r_b$ and $r_c$ similarly. Prove that $r_a + r_b + r_c \geq r$ and find all cases in which equality occurs.

[i]Bosnia - Herzegovina Mathematical Olympiad 2002[/i]
1. **Define the problem and setup:**
   - Let \( \triangle ABC \) be a triangle with an inscribed circle \( G \) of radius \( r \).
   - Let \( r_a \) be the radius of the circle touching \( AB \), \( AC \), and \( G \). Similarly, define \( r_b \) and \( r_c \).

2. **Draw the tangent parallel to \( BC \):**
   - Consider the incircle \( \omega \) of \( \triangle ABC \) and draw the tangent parallel to \( BC \) that touches \( AB \) at \( B' \) and \( AC \) at \( C' \).
   - Let \( r_a' \) be the radius of the incircle \( \omega_a \) of \( \triangle AB'C' \). Similarly, define \( r_b' \) and \( r_c' \).

3. **Claim \( r_a' \le r_a \):**
   - Since line \( B'C' \) separates \( \omega_a \) and \( \omega \), the two circles cannot intersect more than once. This implies \( r_a' \le r_a \).

4. **Establish the inequality:**
   - We need to show that \( r_a + r_b + r_c \geq r \).
   - Using the similarity argument, we have:
     \[
     \frac{r_a}{r} \ge \frac{r_a'}{r} = \frac{r}{r'} = \frac{\frac{K}{s}}{\frac{K}{s-a}} = \frac{s-a}{s}
     \]
     where \( r' \) is the radius of the excircle opposite \( A \), \( K \) is the area of \( \triangle ABC \), and \( s \) is the semiperimeter of \( \triangle ABC \).

5. **Summing up the inequalities:**
   - Summing up the inequalities for \( r_a \), \( r_b \), and \( r_c \), we get:
     \[
     \frac{r_a}{r} + \frac{r_b}{r} + \frac{r_c}{r} \ge \frac{s-a}{s} + \frac{s-b}{s} + \frac{s-c}{s}
     \]
   - Simplifying the right-hand side, we have:
     \[
     \frac{s-a}{s} + \frac{s-b}{s} + \frac{s-c}{s} = \frac{(s-a) + (s-b) + (s-c)}{s} = \frac{3s - (a+b+c)}{s} = \frac{3s - 2s}{s} = 1
     \]
   - Therefore:
     \[
     \frac{r_a}{r} + \frac{r_b}{r} + \frac{r_c}{r} \ge 1
     \]
   - Multiplying both sides by \( r \), we get:
     \[
     r_a + r_b + r_c \ge r
     \]

6. **Equality condition:**
   - Equality holds if and only if \( r_a = r_a' \), \( r_b = r_b' \), and \( r_c = r_c' \).
   - This occurs when \( \triangle ABC \) is equilateral, as in this case, the incircle and the excircles are all congruent.

\(\blacksquare\)

The final answer is \( \boxed{ r_a + r_b + r_c \geq r } \) and equality occurs when \( \triangle ABC \) is equilateral.