To prove that \(\frac{a}{b} + \frac{b}{a} \geq 2\), let's use the **Arithmetic Mean-Geometric Mean (AM-GM) Inequality**. 

### Proof:

1. **Statement of the AM-GM Inequality:**
   For any two positive real numbers \(x\) and \(y\), the AM-GM inequality states:
   \[
   \frac{x + y}{2} \geq \sqrt{xy}.
   \]

2. **Apply AM-GM to \(\frac{a}{b}\) and \(\frac{b}{a}\):**
   Let \(x = \frac{a}{b}\) and \(y = \frac{b}{a}\). Then:
   \[
   \frac{\frac{a}{b} + \frac{b}{a}}{2} \geq \sqrt{\frac{a}{b} \cdot \frac{b}{a}}.
   \]

3. **Simplify the geometric mean:**
   \[
   \sqrt{\frac{a}{b} \cdot \frac{b}{a}} = \sqrt{1} = 1.
   \]

4. **Multiply through by 2 to remove the denominator:**
   \[
   \frac{a}{b} + \frac{b}{a} \geq 2.
   \]

### Conclusion:
\[
\boxed{\frac{a}{b} + \frac{b}{a} \geq 2, \text{ for all } a, b > 0.}
\]

This completes the proof. Equality holds if and only if \( \frac{a}{b} = \frac{b}{a} \), which implies \(a = b\). 

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Would you like additional details or further examples? 

### Related Questions:
1. How does the AM-GM inequality work for three variables?
2. What are the conditions for equality in AM-GM?
3. How can this proof be extended to \(a^2 + b^2 \geq 2ab\)?
4. Can we use calculus to prove this inequality?
5. How does this inequality relate to symmetric functions?

### Tip:
The AM-GM inequality is a powerful tool in inequality proofs and often provides insight into optimization problems. Always check if equality conditions apply to verify your solutions!

Learn how to prove that a/b + b/a is greater than or equal to 2 using the Arithmetic Mean-Geometric Mean (AM-GM) Inequality. This solution provides a step-by-step proof and explores key algebraic concepts, ideal for high school students (Grades 9-12).

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