The value of `(1)/(log_(2)pi)+(1)/(log_(6)pi)` is greater than `2` .

Answer

To solve the inequality \(\frac{1}{\log_{2} \pi} + \frac{1}{\log_{6} \pi} > 2\), we can use properties of logarithms.

### Step-by-Step Solution:

1. **Use the Change of Base Formula**:
We know that \(\frac{1}{\log_{b} a} = \log_{a} b\). Therefore, we can rewrite the terms:
\[ \frac{1}{\log_{2} \pi} = \log_{\pi} 2 \quad \text{and} \quad \frac{1}{\log_{6} \pi} = \log_{\pi} 6 \]
Thus, the inequality becomes:
\[ \log_{\pi} 2 + \log_{\pi} 6 > 2 \]

2. **Combine the Logarithms**:
Using the property of logarithms that states \(\log_{a} b + \log_{a} c = \log_{a} (bc)\), we can combine the logarithms:
\[ \log_{\pi} (2 \cdot 6) > 2 \]
This simplifies to:
\[ \log_{\pi} 12 > 2 \]

3. **Rewrite the Logarithmic Inequality**:
The inequality \(\log_{\pi} 12 > 2\) can be rewritten in exponential form:
\[ 12 > \pi^2 \]

4. **Evaluate the Inequality**:
We need to check if \(12 > \pi^2\). We know that \(\pi \approx 3.14\), thus:
\[ \pi^2 \approx 3.14^2 \approx 9.8596 \]
Since \(12 > 9.8596\), the inequality holds true.

### Conclusion:
Therefore, we conclude that:
\[ \frac{1}{\log_{2} \pi} + \frac{1}{\log_{6} \pi} > 2 \]