If `x_(1)` and `x_(2)` are positive quantities, then the condition for the difference between the arithmetic mean and the geometric mean to be greater than I is:

Options

A. `x_(1)+x_(2)gt2sqrt(x_(1)x_(2))`
B. `sqrt(x_(1))+sqrt(x_(2))gtsqrt(2)`
C. `|sqrt(x_(1))-sqrt(x_(2))|gtsqrt(2)`
D. `x_(1)+x_(2)lt2(sqrt(x_(1)+x_(2))+1)`

Answer

To solve the problem, we need to find the condition under which the difference between the arithmetic mean (AM) and the geometric mean (GM) of two positive quantities \( x_1 \) and \( x_2 \) is greater than 1.

### Step-by-Step Solution:

1. **Define Arithmetic Mean and Geometric Mean**:
- The arithmetic mean (AM) of \( x_1 \) and \( x_2 \) is given by:
\[ AM = \frac{x_1 + x_2}{2} \]
- The geometric mean (GM) of \( x_1 \) and \( x_2 \) is given by:
\[ GM = \sqrt{x_1 \cdot x_2} \]

2. **Set Up the Inequality**:
- We want to find the condition such that:
\[ AM - GM > 1 \]
- Substituting the expressions for AM and GM, we have:
\[ \frac{x_1 + x_2}{2} - \sqrt{x_1 \cdot x_2} > 1 \]

3. **Multiply Through by 2**:
- To eliminate the fraction, multiply both sides of the inequality by 2:
\[ x_1 + x_2 - 2\sqrt{x_1 \cdot x_2} > 2 \]

4. **Rearrange the Inequality**:
- Rearranging gives us:
\[ x_1 + x_2 - 2\sqrt{x_1 \cdot x_2} - 2 > 0 \]

5. **Recognize the Perfect Square**:
- Notice that \( x_1 + x_2 - 2\sqrt{x_1 \cdot x_2} \) can be rewritten as:
\[ (\sqrt{x_1} - \sqrt{x_2})^2 \]
- Therefore, we can rewrite the inequality as:
\[ (\sqrt{x_1} - \sqrt{x_2})^2 > 2 \]

6. **Taking the Square Root**:
- Taking the square root of both sides (since both \( x_1 \) and \( x_2 \) are positive, the square root is valid):
\[ |\sqrt{x_1} - \sqrt{x_2}| > \sqrt{2} \]

### Final Condition:
Thus, the condition for the difference between the arithmetic mean and the geometric mean to be greater than 1 is:
\[ |\sqrt{x_1} - \sqrt{x_2}| > \sqrt{2} \]