Đáp án+Giải thích các bước giải:
\[a,b,c \in N^{*}\\\to a,b,c>0\\\dfrac{a}{b}+\dfrac{b}{a} \geq 2\\\to \Bigg(\sqrt{\dfrac{a}{b}}\Bigg)^{2}-2.\Bigg(\sqrt{\dfrac{a}{b}}\Bigg).\Bigg(\sqrt{\dfrac{b}{a}}\Bigg)+\Bigg(\sqrt{\dfrac{b}{a}}\Bigg)^{2} \geq 0\\\to \Bigg(\sqrt{\dfrac{a}{b}}-\sqrt{\dfrac{b}{a}}\Bigg)^{2} \geq 0(\text{luôn đúng})\\\text{Dấu "=" xảy ra khi a=b}\]
Hoàn toàn tương tự:
\[S=\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b} \geq 6\]