Ifa, b, c are positive real number such that `ab^(2)c^(3) = 64` then minimum value of `((1)/(a) + (2)/(b) + (3)/(c))` is equal to

Options

A. 6
B. 2
C. 3
D. None of these

Answer

To find the minimum value of the expression \( \frac{1}{a} + \frac{2}{b} + \frac{3}{c} \) given the constraint \( ab^2c^3 = 64 \), we can use the method of inequalities, specifically the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Set Up the Problem**:
We are given that \( ab^2c^3 = 64 \). We need to minimize \( \frac{1}{a} + \frac{2}{b} + \frac{3}{c} \).

2. **Express the Terms**:
We can express the terms in the sum as follows:
\[ \frac{1}{a}, \quad \frac{1}{b}, \quad \frac{1}{b}, \quad \frac{1}{c}, \quad \frac{1}{c}, \quad \frac{1}{c} \]
This gives us a total of 6 terms.

3. **Apply AM-GM Inequality**:
According to the AM-GM inequality, the arithmetic mean of non-negative numbers is greater than or equal to the geometric mean. Therefore, we have:
\[ \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{b} + \frac{1}{c} + \frac{1}{c} + \frac{1}{c}}{6} \geq \sqrt[6]{\frac{1}{a} \cdot \frac{1}{b^2} \cdot \frac{1}{c^3}} \]

4. **Simplify the Right Side**:
The right side simplifies to:
\[ \sqrt[6]{\frac{1}{ab^2c^3}} = \frac{1}{(ab^2c^3)^{1/6}} \]
Since we know \( ab^2c^3 = 64 \), we can substitute this in:
\[ (ab^2c^3)^{1/6} = 64^{1/6} = 2 \]
Therefore, we have:
\[ \frac{\frac{1}{a} + \frac{2}{b} + \frac{3}{c}}{6} \geq \frac{1}{2} \]

5. **Multiply by 6**:
Multiplying both sides by 6 gives:
\[ \frac{1}{a} + \frac{2}{b} + \frac{3}{c} \geq 3 \]

6. **Conclusion**:
The minimum value of \( \frac{1}{a} + \frac{2}{b} + \frac{3}{c} \) is \( 3 \).

### Final Answer:
The minimum value of \( \frac{1}{a} + \frac{2}{b} + \frac{3}{c} \) is \( 3 \).