In any triangle, the minimum value of `r_(1) r_(2) r_(3) //r^(3)` is equal to

Options

A. 1
B. 9
C. 27
D. none of these

Explanation

To solve the problem of finding the minimum value of \( \frac{r_1 r_2 r_3}{r^3} \) in any triangle, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Variables**:
- Let \( r_1, r_2, r_3 \) be the inradii of the triangle's excenters opposite to vertices \( A, B, C \) respectively.
- Let \( r \) be the inradius of the triangle.
- We need to find the minimum value of \( \frac{r_1 r_2 r_3}{r^3} \).

2. **Using the Inequality**:
- We can use the property that the geometric mean (GM) is greater than or equal to the harmonic mean (HM). This gives us:
\[ \sqrt[3]{\frac{r_1}{r} \cdot \frac{r_2}{r} \cdot \frac{r_3}{r}} \geq \frac{3}{\frac{r}{r_1} + \frac{r}{r_2} + \frac{r}{r_3}} \]

3. **Substituting Values**:
- We know that:
\[ \frac{r}{r_1} = \frac{s-a}{s}, \quad \frac{r}{r_2} = \frac{s-b}{s}, \quad \frac{r}{r_3} = \frac{s-c}{s} \]
- Here, \( s \) is the semi-perimeter of the triangle.

4. **Simplifying the Expression**:
- Substitute these values into the inequality:
\[ \frac{r_1}{r} \cdot \frac{r_2}{r} \cdot \frac{r_3}{r} \geq \left( \frac{3}{\frac{s-a}{s} + \frac{s-b}{s} + \frac{s-c}{s}} \right)^3 \]
- This simplifies to:
\[ \frac{r_1 r_2 r_3}{r^3} \geq \frac{27s^3}{(s-a + s-b + s-c)^3} \]

5. **Finding the Minimum Value**:
- The expression \( s-a + s-b + s-c = 3s \), thus:
\[ \frac{27s^3}{(3s)^3} = \frac{27s^3}{27s^3} = 1 \]
- Therefore, we conclude that:
\[ \frac{r_1 r_2 r_3}{r^3} \geq 27 \]

6. **Conclusion**:
- The minimum value of \( \frac{r_1 r_2 r_3}{r^3} \) is \( 27 \).

### Final Answer:
The minimum value of \( \frac{r_1 r_2 r_3}{r^3} \) is **27**.