\[P=(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})\] \[\to P=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\] \[\to P=3+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}\] \[\to P=3+\Bigg(\dfrac{a}{b}+\dfrac{b}{a}\Bigg)+\Bigg(\dfrac{a}{c}+\dfrac{c}{a}\Bigg)+\Bigg(\dfrac{b}{c}+\dfrac{c}{b}\Bigg)\ge 3+2+2+2=9\]
Dấu $"="$ xảy ra $⇔a=b=c$
