a) $\dfrac{a}{b}+\dfrac{b}{a} ≥ 2\sqrt[]{\dfrac{a}{b}.\dfrac{b}{a}} = 2$
Dấu "=" xảy ra $⇔a=b$
b) $\dfrac{1}{a}+\dfrac{1}{b} ≥ 2\sqrt[]{\dfrac{1}{ab}}$
$a+b ≥ 2\sqrt[]{ab}$
Nên $\bigg(\dfrac{1}{a}+\dfrac{1}{b}\bigg).(a+b) ≥ 4$
$\to \dfrac{1}{a}+\dfrac{1}{b} ≥ \dfrac{4}{a+b}$