If three positive real numbers a,b,c are in AP such that abc=4, then the minimum value of b is

Options

A. `2^(1//3)`
B. `2^(2//3)`
C. `2^(1//2)`
D. `2^(3//2)`

Explanation

To find the minimum value of \( b \) given that \( a, b, c \) are in arithmetic progression (AP) and their product \( abc = 4 \), we can follow these steps:

### Step 1: Express \( a \) and \( c \) in terms of \( b \)
Since \( a, b, c \) are in AP, we can express \( a \) and \( c \) in terms of \( b \) and a common difference \( d \):
\[ a = b - d \quad \text{and} \quad c = b + d \]

### Step 2: Write the product \( abc \)
Substituting the expressions for \( a \) and \( c \) into the product \( abc \):
\[ abc = (b - d) \cdot b \cdot (b + d) \]
This simplifies to:
\[ abc = b(b^2 - d^2) = b^3 - bd^2 \]

### Step 3: Set the product equal to 4
Since we know \( abc = 4 \), we can set up the equation:
\[ b^3 - bd^2 = 4 \]
Rearranging gives:
\[ b^3 - 4 = bd^2 \]

### Step 4: Solve for \( d^2 \)
From the equation \( b^3 - 4 = bd^2 \), we can express \( d^2 \):
\[ d^2 = \frac{b^3 - 4}{b} \]

### Step 5: Minimize \( b \)
To find the minimum value of \( b \), we need to ensure that \( d^2 \) is non-negative (since \( d^2 \geq 0 \)):
\[ \frac{b^3 - 4}{b} \geq 0 \]
This implies:
\[ b^3 - 4 \geq 0 \quad \Rightarrow \quad b^3 \geq 4 \quad \Rightarrow \quad b \geq \sqrt[3]{4} \]

### Step 6: Calculate the minimum value of \( b \)
The minimum value of \( b \) is:
\[ b = \sqrt[3]{4} = 2^{2/3} \]

### Conclusion
Thus, the minimum value of \( b \) is \( 2^{2/3} \).