Theo suy luận thì nó như thế này
$(a+b+c) (\dfrac{1}{a} + \dfrac{1}{b} +\dfrac{1}{c} )$ $\geq$ $9$
=> $\dfrac{a}{a} + \dfrac{a}{b} + \dfrac{a}{c} + \dfrac{b}{a}+\dfrac{b}{c}+\dfrac{b}{b} + \dfrac{c}{c}$
$+ \dfrac{c}{a} + \dfrac{c}{b}$ $\geq$ $9$
$=> 3 + ( \dfrac{a}{b} + \dfrac{b}{a}) + (\dfrac{b}{c} + \dfrac{c}{b}) + ( \dfrac{a}{c} + \dfrac{c}{a})$ $\geq$ $9$
Mà theo Cauchy ta có
$\dfrac{a}{b} + \dfrac{b}{a} \geq 2 \sqrt[]{\dfrac{a}{b} . \dfrac{b}{a}}$ $= 2$
Tương tự
$\dfrac{a}{c} + \dfrac{c}{a} \geq 2 \sqrt[]{\dfrac{a}{c} . \dfrac{c}{a}}$ $= 2$
$\dfrac{b}{c} + \dfrac{c}{b} \geq 2 \sqrt[]{\dfrac{b}{c} . \dfrac{c}{b}}$ $= 2$
$=> (a+b+c) (\dfrac{1}{a} + \dfrac{1}{b} +\dfrac{1}{c} )$ $\geq$ $3+2+2+2= 9$
Dấu $"="$ khi $a = b = c$
