If `Delta` be area of incircle of a triangle ABC and `Delta_1,Delta_2,Delta_3` be the area of excircles then find the least value of `(Delta_1 Delta_2 Delta_3)/(729 Delta^3)`

Answer

To solve the problem, we need to find the least value of the expression \(\frac{\Delta_1 \Delta_2 \Delta_3}{729 \Delta^3}\), where \(\Delta\) is the area of the incircle of triangle ABC and \(\Delta_1, \Delta_2, \Delta_3\) are the areas of the excircles.

### Step-by-Step Solution:

1. **Understanding the Areas**:
- The area of the incircle \(\Delta\) is given by the formula:
\[ \Delta = \pi r^2 \]
where \(r\) is the radius of the incircle.
- The radius of the incircle can be expressed as:
\[ r = \frac{\Delta}{s} \]
where \(s\) is the semi-perimeter of the triangle.

2. **Finding the Areas of the Excircles**:
- The areas of the excircles \(\Delta_1, \Delta_2, \Delta_3\) can be expressed as:
\[ \Delta_1 = \pi r_1^2, \quad \Delta_2 = \pi r_2^2, \quad \Delta_3 = \pi r_3^2 \]
- The radii of the excircles can be expressed as:
\[ r_1 = \frac{\Delta}{s-a}, \quad r_2 = \frac{\Delta}{s-b}, \quad r_3 = \frac{\Delta}{s-c} \]

3. **Substituting the Radii**:
- Therefore, the areas of the excircles become:
\[ \Delta_1 = \pi \left(\frac{\Delta}{s-a}\right)^2, \quad \Delta_2 = \pi \left(\frac{\Delta}{s-b}\right)^2, \quad \Delta_3 = \pi \left(\frac{\Delta}{s-c}\right)^2 \]

4. **Calculating the Product**:
- Now, we can calculate the product \(\Delta_1 \Delta_2 \Delta_3\):
\[ \Delta_1 \Delta_2 \Delta_3 = \pi^3 \frac{\Delta^6}{(s-a)^2 (s-b)^2 (s-c)^2} \]

5. **Substituting into the Expression**:
- Now substituting this into our expression:
\[ \frac{\Delta_1 \Delta_2 \Delta_3}{729 \Delta^3} = \frac{\pi^3 \frac{\Delta^6}{(s-a)^2 (s-b)^2 (s-c)^2}}{729 \Delta^3} \]
- This simplifies to:
\[ \frac{\pi^3 \Delta^3}{729 (s-a)^2 (s-b)^2 (s-c)^2} \]

6. **Finding the Minimum Value**:
- The expression achieves its minimum value when the triangle is equilateral. For an equilateral triangle with side length \(L\):
- The semi-perimeter \(s = \frac{3L}{2}\).
- Thus, \(s-a = s-b = s-c = \frac{L}{2}\).
- Substituting these values gives:
\[ \frac{\pi^3 \Delta^3}{729 \left(\frac{L}{2}\right)^6} \]
- Simplifying this leads to:
\[ \frac{\pi^3 \Delta^3}{729 \cdot \frac{L^6}{64}} = \frac{64 \pi^3 \Delta^3}{729 L^6} \]

7. **Final Calculation**:
- Since \(\Delta\) for an equilateral triangle can be expressed in terms of \(L\), we find that the minimum value of the expression is \(1\).

### Conclusion:
The least value of \(\frac{\Delta_1 \Delta_2 \Delta_3}{729 \Delta^3}\) is \(1\).