Đáp án + Giải thích các bước giải:
`@` $\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{a+c}{b}=\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{a}{b}+\dfrac{c}{b}$
$=\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)$
Áp dụng bất đẳng thức Cô-si ($a+b≥2\sqrt{ab}$) ta có:
$\dfrac{a}{c}+\dfrac{c}{a}≥2\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=2$
$\dfrac{b}{c}+\dfrac{c}{b}≥2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}=2$
$\dfrac{b}{a}+\dfrac{a}{b}≥2\sqrt{\dfrac{b}{a}.\dfrac{a}{b}}=2$
$→ \left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{b}{a}+\dfrac{a}{b}\right) ≥ 2+2+2=6$ (đpcm)
Dấu $"="$ xảy ra khi $a=b=c$
