If `a,b,c,d in R^(+)-{1}`, then the minimum value of `log_(d) a+ log_(c)b+log _(a)c+log_(b)d` is

Options

A. 4
B. 2
C. 1
D. none of these

Explanation

To find the minimum value of the expression \( \log_d a + \log_c b + \log_a c + \log_b d \) for positive real numbers \( a, b, c, d \) not equal to 1, we can utilize the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-step Solution:

1. **Rewrite the Expression**:
We start with the expression:
\[ E = \log_d a + \log_c b + \log_a c + \log_b d \]

2. **Apply Change of Base Formula**:
Using the change of base formula for logarithms, we can rewrite each term:
\[ \log_d a = \frac{\log a}{\log d}, \quad \log_c b = \frac{\log b}{\log c}, \quad \log_a c = \frac{\log c}{\log a}, \quad \log_b d = \frac{\log d}{\log b} \]
Therefore, the expression becomes:
\[ E = \frac{\log a}{\log d} + \frac{\log b}{\log c} + \frac{\log c}{\log a} + \frac{\log d}{\log b} \]

3. **Use AM-GM Inequality**:
By applying the AM-GM inequality, we have:
\[ \frac{\log a}{\log d} + \frac{\log b}{\log c} + \frac{\log c}{\log a} + \frac{\log d}{\log b} \geq 4 \sqrt[4]{\frac{\log a}{\log d} \cdot \frac{\log b}{\log c} \cdot \frac{\log c}{\log a} \cdot \frac{\log d}{\log b}} \]

4. **Simplify the Geometric Mean**:
Notice that the product inside the fourth root simplifies:
\[ \frac{\log a}{\log d} \cdot \frac{\log b}{\log c} \cdot \frac{\log c}{\log a} \cdot \frac{\log d}{\log b} = 1 \]
Thus, we have:
\[ \sqrt[4]{1} = 1 \]

5. **Conclude the Inequality**:
Therefore, we conclude:
\[ E \geq 4 \cdot 1 = 4 \]

6. **Minimum Value**:
The minimum value of \( E \) is \( 4 \).

### Final Answer:
The minimum value of \( \log_d a + \log_c b + \log_a c + \log_b d \) is \( 4 \).