VT=1+\(\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\)
=\(3\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)
\(\ge\) 3+2+2+2=9 (áp dụng BĐT \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\))ĐPCM