We are tasked with solving the inequality \( 1 + \frac{1}{2x} > 1 + x \) where \( x > 0 \).

### Step-by-Step Solution:

1. **Simplify both sides**:
   \[
   1 + \frac{1}{2x} > 1 + x
   \]
   Subtract 1 from both sides:
   \[
   \frac{1}{2x} > x
   \]

2. **Eliminate the fraction** by multiplying both sides by \( 2x \), keeping in mind that \( x > 0 \), so multiplying by \( 2x \) will not change the direction of the inequality:
   \[
   1 > 2x^2
   \]

3. **Solve for \( x \)**:
   Divide both sides by 2:
   \[
   \frac{1}{2} > x^2
   \]
   Taking the square root of both sides (since \( x > 0 \), we only take the positive root):
   \[
   x < \frac{1}{\sqrt{2}}
   \]

### Final Answer:
The solution is:
\[
0 < x < \frac{1}{\sqrt{2}}
\]

This means that for the inequality \( 1 + \frac{1}{2x} > 1 + x \) to hold, \( x \) must lie between 0 and \( \frac{1}{\sqrt{2}} \).

Would you like more details on any step or have any questions?

### Related Questions:
1. What happens if \( x \leq 0 \) in the original inequality?
2. How would the solution change if the inequality were \( 1 + \frac{1}{2x} \geq 1 + x \)?
3. Can you express \( \frac{1}{\sqrt{2}} \) as a decimal?
4. How does this inequality behave as \( x \to 0^+ \)?
5. What if the inequality were reversed, i.e., \( 1 + \frac{1}{2x} < 1 + x \)?

### Tip:
Always consider the domain restrictions when working with inequalities that involve variables in the denominator.

Learn how to solve the inequality 1 + 1/2x > 1 + x for x > 0 with a detailed step-by-step guide. This algebra problem involves inequalities, eliminating fractions, and solving quadratic equations. Suitable for students in Grades 9-11.

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