`|b_(xy)+ b_(yx)|ge`_________

Answer

To solve the question regarding the relationship between the regression coefficients \( b_{xy} \) and \( b_{yx} \), we will follow these steps:

### Step-by-Step Solution:

1. **Understanding Regression Coefficients**:
- Let \( b_{xy} \) be the regression coefficient of \( x \) on \( y \).
- Let \( b_{yx} \) be the regression coefficient of \( y \) on \( x \).
- We denote the correlation coefficient as \( r \).

2. **Using the Relationship between Correlation and Regression Coefficients**:
- There is a known relationship which states that:
\[ r^2 = b_{xy} \cdot b_{yx} \]
- From this, we can express \( r \) as:
\[ r = \pm \sqrt{b_{xy} \cdot b_{yx}} \]

3. **Applying the Absolute Value**:
- Taking the absolute value, we have:
\[ |r| = \sqrt{|b_{xy} \cdot b_{yx}|} \]

4. **Using the Arithmetic Mean-Geometric Mean Inequality**:
- By the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ \frac{b_{xy} + b_{yx}}{2} \geq \sqrt{b_{xy} \cdot b_{yx}} \]
- This implies:
\[ |b_{xy} + b_{yx}| \geq 2\sqrt{|b_{xy} \cdot b_{yx}|} \]

5. **Substituting the Expression for \( |r| \)**:
- Since \( |r| = \sqrt{|b_{xy} \cdot b_{yx}|} \), we can substitute this into our inequality:
\[ |b_{xy} + b_{yx}| \geq 2|r| \]

### Final Answer:
Thus, we conclude that:
\[ |b_{xy} + b_{yx}| \geq 2|r| \]