South African Mathematics Olympiad Second Round

It is easy to see that if $a=b$, then the expression is equal to $2$, and a few trials will suggest that $2$ is the smallest value. A proof is that

Let $x=\frac{a}{b}$, so $x>0$. Then $\frac{a}{b}+\frac{b}{a}=x+\frac{1}{x}$.

By the AM-GM inequality:

$$x+\frac{1}{x}\ge 2\sqrt{x\cdot \frac{1}{x}}=2\sqrt{1}=2.$$

Equality holds when $x=1$, i.e., $a=b$.

Therefore, the smallest possible value is $2$.