Given `agt0,bgt0,agtb` and `c!=0`. Which in-equality is not always correct?

Options

A. `a+cgt+c`
B. `a-cgtb-c`
C. `acgtbc`
D. `a/(c^(2))gtb/(c^(2))`

Answer

To solve the problem, we need to analyze the given inequalities and determine which one is not always correct based on the conditions provided: \( a > 0 \), \( b > 0 \), \( a > b \), and \( c \neq 0 \).

### Step-by-Step Solution:

1. **Understanding the Conditions**:
- We have \( a > 0 \)
- \( b > 0 \)
- \( a > b \)
- \( c \neq 0 \) (which means \( c \) can be either positive or negative)

2. **Analyzing Each Option**:
- **Option 1**: \( a + c > b + c \)
- Rearranging gives \( a > b \), which is true as per the given conditions.
- **Option 2**: \( a - c > b - c \)
- Rearranging gives \( a > b \), which is also true as per the given conditions.
- **Option 3**: \( ac > bc \)
- This depends on the sign of \( c \):
- If \( c > 0 \), then \( ac > bc \) is true because \( a > b \).
- If \( c < 0 \), then multiplying both sides of \( a > b \) by \( c \) reverses the inequality, leading to \( ac < bc \). Thus, this inequality is not always true.
- **Option 4**: \( \frac{a}{c^2} > \frac{b}{c^2} \)
- Since \( c^2 > 0 \) (as it is the square of any non-zero number), we can divide both sides by \( c^2 \) without changing the inequality. Therefore, \( a > b \) implies \( \frac{a}{c^2} > \frac{b}{c^2} \) is always true.

3. **Conclusion**:
- The inequality that is not always correct is **Option 3: \( ac > bc \)**.

### Final Answer:
The inequality that is not always correct is \( ac > bc \).