$$\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}=2\cdot \frac{a^2+b^2}{2}\cdot \frac{1}{ab}\ldots (1)$$
$$ab=\sqrt{a^2b^2}\leq \frac{a^2+b^2}{2}\ldots (2)$$
$$(1),(2)\Rightarrow \frac{a}{b}+\frac{b}{a}\geq 2\cdot ab\cdot \frac{1}{ab}=2$$ veya
$$\sqrt{\frac{a}{b}\cdot\frac{b}{a}}\leq \frac{\frac{a}{b}+\frac{b}{a}}{2}\Rightarrow 2\leq \frac{a}{b}+\frac{b}{a}$$