If x and y are two distinct positive integers, then mean of x and y is always greater than______

Options

A. `xy`
B. `sqrt2 xy`
C. `2 sqrtxy`
D. `sqrt (x y)`

Answer

To solve the problem, we need to find out what the mean of two distinct positive integers \( x \) and \( y \) is always greater than.

### Step-by-Step Solution:

1. **Understanding the Mean**:
The mean (average) of two numbers \( x \) and \( y \) is calculated using the formula:
\[ \text{Mean} = \frac{x + y}{2} \]

2. **Identifying Distinct Positive Integers**:
Since \( x \) and \( y \) are distinct positive integers, we know that \( x \neq y \) and both \( x \) and \( y \) are greater than 0.

3. **Using the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality)**:
The AM-GM inequality states that for any two non-negative numbers \( a \) and \( b \):
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
In our case, we can apply this to \( x \) and \( y \):
\[ \frac{x + y}{2} \geq \sqrt{xy} \]

4. **Conclusion**:
From the AM-GM inequality, we can conclude that the mean of \( x \) and \( y \) is always greater than or equal to the geometric mean of \( x \) and \( y \):
\[ \frac{x + y}{2} \geq \sqrt{xy} \]
Therefore, the mean of \( x \) and \( y \) is always greater than \( \sqrt{xy} \).

### Final Answer:
The mean of \( x \) and \( y \) is always greater than \( \sqrt{xy} \).