Ta có
$ (a+b+c) (\dfrac{1}{a} + \dfrac{1}{b} +\dfrac{1}{c} ) $
$ = \dfrac{a}{a} + \dfrac{b}{b} + \dfrac{c}{c} + \dfrac{a}{b} + \dfrac{a}{c} + \dfrac{b}{c} + \dfrac{b}{a}$$+ \dfrac{c}{b} + \dfrac{c}{a} $
$ = 3 + ( \dfrac{a}{b} + \dfrac{b}{a}) + ( \dfrac{a}{c} + \dfrac{c}{a}) + ( \dfrac{b}{c} + \dfrac{c}{b} ) $
Áp dụng bđt Cauchy (với $a,b,c > 0$) ta có
$ \dfrac{a}{b} + \dfrac{b}{a} \geq 2 \sqrt[]{\dfrac{a}{b} . \dfrac{b}{a}} = 2 $
Tương tự ta có
$ \dfrac{a}{c} + \dfrac{c}{a} \geq 2 \sqrt[]{\dfrac{a}{c} . \dfrac{c}{a}} = 2 $
$ \dfrac{c}{b} + \dfrac{b}{c} \geq 2 \sqrt[]{\dfrac{b}{c} . \dfrac{c}{b}} = 2 $
=> $ (a+b+c) (\dfrac{1}{a} + \dfrac{1}{b} +\dfrac{1}{c} ) \geq 3 + 2 + 2 + 2 = 9 $
Dấu $"="$ xảy ra khi $a = b = c$
