In any triangle ABC, find the least value of `(r_(1) + r_(2) + r_(3))/(r)`

Explanation

To find the least value of \((r_1 + r_2 + r_3)/r\) in any triangle \(ABC\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Terms**:
- Let \(r_1\), \(r_2\), and \(r_3\) be the inradii of triangles formed by the vertices of triangle \(ABC\) and the sides opposite to these vertices.
- Let \(r\) be the inradius of triangle \(ABC\).

2. **Using the Relation**:
- We know that:
\[ \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{s-a}{\Delta} + \frac{s-b}{\Delta} + \frac{s-c}{\Delta} \]
where \(s\) is the semi-perimeter \((s = \frac{a + b + c}{2})\) and \(\Delta\) is the area of triangle \(ABC\).

3. **Finding a Common Denominator**:
- Taking the common denominator \(\Delta\):
\[ \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{(s-a) + (s-b) + (s-c)}{\Delta} \]
- Simplifying the numerator:
\[ (s-a) + (s-b) + (s-c) = 3s - (a + b + c) = 3s - 2s = s \]
- Thus, we have:
\[ \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{s}{\Delta} \]

4. **Relating to Inradius**:
- Recall that the inradius \(r\) is given by:
\[ r = \frac{\Delta}{s} \]
- Therefore, we can write:
\[ \frac{s}{\Delta} = \frac{1}{r} \]
- This leads to:
\[ \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r} \]

5. **Applying the AM-HM Inequality**:
- By the Arithmetic Mean - Harmonic Mean (AM-HM) inequality:
\[ \frac{r_1 + r_2 + r_3}{3} \geq \frac{3}{\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}} \]
- Substituting \(\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}\):
\[ \frac{r_1 + r_2 + r_3}{3} \geq 3r \]
- This implies:
\[ r_1 + r_2 + r_3 \geq 9r \]

6. **Final Result**:
- Dividing both sides by \(r\):
\[ \frac{r_1 + r_2 + r_3}{r} \geq 9 \]
- Therefore, the least value of \(\frac{r_1 + r_2 + r_3}{r}\) is \(9\).

### Conclusion:
The least value of \(\frac{r_1 + r_2 + r_3}{r}\) in any triangle \(ABC\) is \(9\).