Giải thích các bước giải:
Ta có :
$\dfrac{a}{bc}+\dfrac{b}{ca}\ge 2\sqrt{\dfrac{a}{bc}.\dfrac{b}{ca}}=2.\dfrac{1}{c}$
Chứng minh tương tự :
$\dfrac{b}{ac}+\dfrac{c}{ab}\ge 2.\dfrac{1}{a}$
$\dfrac{c}{ab}+\dfrac{a}{bc}\ge 2.\dfrac{1}{b}$
Cộng vế với vế
$\rightarrow \dfrac{a}{bc}+\dfrac{b}{ca}+\dfrac{b}{ac}+\dfrac{c}{ab}+\dfrac{c}{ab}+\dfrac{a}{bc}\ge 2.\dfrac{1}{c}+2.\dfrac{1}{a}+2.\dfrac{1}{b} $
$\rightarrow 2(\dfrac{a}{bc}+\dfrac{b}{ac}+\dfrac{c}{ab})\ge 2.(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}) $
$\rightarrow \dfrac{a}{bc}+\dfrac{b}{ac}+\dfrac{c}{ab}\ge \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} $
