Bài 4.8 trang 104 SBT đại số 10

\((a + b + c)(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}) = 1 + 1 + 1 + (\dfrac{a}{b} + \dfrac{b}{a}) + (\dfrac{a}{c} + \dfrac{c}{a}) + (\dfrac{b}{c} + \dfrac{c}{b})\)

\( \ge 3 + 2 + 2 + 2 = 9\)\( \Leftrightarrow \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \ge \dfrac{9}{{a + b + c}}\)