Giải thích các bước giải:
Vì $a,b,c$ dương nên theo BĐT AM - GM ta có :
$\dfrac{ab}{c}+\dfrac{ac}{b} ≥ 2\sqrt[]{\dfrac{ab}{c}.\dfrac{ac}{b}} = 2a$
$\dfrac{ab}{c}+\dfrac{cb}{a} ≥ 2\sqrt[]{\dfrac{ab}{c}.\dfrac{bc}{a}} = 2b$
$\dfrac{cb}{a}+\dfrac{ac}{b} ≥ 2\sqrt[]{\dfrac{cb}{a}.\dfrac{ac}{b}} = 2c$
$\to 2.\bigg(\dfrac{ab}{c}+\dfrac{ac}{b} +\dfrac{bc}{a}\bigg) ≥ 2.(a+b+c)$
$\to \dfrac{ab}{c}+\dfrac{ac}{b} +\dfrac{bc}{a}≥ a+b+c$
Dấu "=" xảy ra $⇔a=b=c$
