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 1 is

Options

A. `x_(1)+x_(2)gt2sqrt(x_(1)x_(2))`
B. `sqrt(x_(1))+sqrt(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)`

Explanation

To solve the problem, we need to find the condition for the difference between the arithmetic mean (AM) and the geometric mean (GM) of two positive quantities \( x_1 \) and \( x_2 \) to be 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:
\[ GM = \sqrt{x_1 x_2} \]

2. **Set Up the Inequality**:
- We need to find when the difference between the AM and GM is greater than 1:
\[ AM - GM > 1 \]
- Substitute the expressions for AM and GM:
\[ \frac{x_1 + x_2}{2} - \sqrt{x_1 x_2} > 1 \]

3. **Rearranging the Inequality**:
- Rearranging gives:
\[ \frac{x_1 + x_2}{2} > \sqrt{x_1 x_2} + 1 \]
- Multiply both sides by 2 (since both \( x_1 \) and \( x_2 \) are positive, this does not change the inequality):
\[ x_1 + x_2 > 2\sqrt{x_1 x_2} + 2 \]

4. **Rearranging Further**:
- Now, we can rearrange this to:
\[ x_1 + x_2 - 2\sqrt{x_1 x_2} > 2 \]

5. **Using the Identity**:
- The left side can be rewritten using the identity \( a + b - 2\sqrt{ab} = (\sqrt{a} - \sqrt{b})^2 \):
\[ (\sqrt{x_1} - \sqrt{x_2})^2 > 2 \]

6. **Taking Square Roots**:
- Taking the square root of both sides (and considering the absolute value since we are dealing with squares):
\[ |\sqrt{x_1} - \sqrt{x_2}| > \sqrt{2} \]

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