Problem: Prove that for all $ (a,b)\in(\mathbb{R}^+)^2$, $\frac{a}{b}+\frac{b}{a}\ge 2$. 1. Let $x=\frac{a}{b}$ which is positive because $a,b>0$. 2. Then $\frac{a}{b}+\frac{b}{a}=x+\frac{1}{x}$. 3. By the arithmetic--geometric mean inequality for positive $x$ we have $x+\frac{1}{x}\ge 2\sqrt{x\cdot\frac{1}{x}}=2$. 4. Alternatively, consider $(x-1)^2\ge 0$ which expands to $x^2-2x+1\ge 0$ and dividing by $x>0$ gives $x+\frac{1}{x}-2\ge 0$, so $x+\frac{1}{x}\ge 2$. 5. Substituting back $x=\frac{a}{b}$ yields $\frac{a}{b}+\frac{b}{a}\ge 2$ with equality iff $x=1$, i.e. $a=b$. Final answer: $\displaystyle \frac{a}{b}+\frac{b}{a}\ge 2$ for all $a,b>0$, with equality when $a=b$.