In any triangle ABC `sum (sin^2A+sinA+1)/sinA` is always greater than or equal

Options

A. 9
B. 3
C. 27
D. None of these

Answer

To solve the problem, we need to show that the expression

\[ \sum \frac{\sin^2 A + \sin A + 1}{\sin A} \]

is always greater than or equal to 9 for any triangle ABC.

### Step-by-Step Solution:

1. **Rewrite the Expression**:
The given expression can be rewritten as:
\[ \sum \left( \sin A + 1 + \frac{1}{\sin A} \right) \]
This means we will evaluate:
\[ \left( \sin A + 1 + \frac{1}{\sin A} \right) + \left( \sin B + 1 + \frac{1}{\sin B} \right) + \left( \sin C + 1 + \frac{1}{\sin C} \right) \]

2. **Focus on One Term**:
Let's analyze one term:
\[ \sin A + 1 + \frac{1}{\sin A} \]
We can denote this term as \( f(\sin A) = \sin A + 1 + \frac{1}{\sin A} \).

3. **Find the Minimum Value**:
To find the minimum value of \( f(\sin A) \), we can use the AM-GM inequality:
\[ \sin A + \frac{1}{\sin A} \geq 2 \]
This is because the arithmetic mean is always greater than or equal to the geometric mean.

4. **Combine the Results**:
Adding 1 to both sides gives:
\[ \sin A + 1 + \frac{1}{\sin A} \geq 2 + 1 = 3 \]

5. **Apply to All Angles**:
Since this holds for angles A, B, and C, we can write:
\[ f(\sin A) + f(\sin B) + f(\sin C) \geq 3 + 3 + 3 = 9 \]

6. **Conclusion**:
Therefore, we conclude that:
\[ \sum \frac{\sin^2 A + \sin A + 1}{\sin A} \geq 9 \]

### Final Result:
Thus, the expression is always greater than or equal to 9.

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