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 $a = b$.

---

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!