\[a+b+c\ge 3\Bigg(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Bigg)\] \[↔a+b+c\ge 3.\dfrac{ab+bc+ca}{abc}\] \[↔(a+b+c)^2\ge 3(ab+bc+ca)\] \[↔a^2+b^2+c^2\ge ab+bc+ca\] \[↔2a^2+2b^2+2c^2\ge 2ab+2bc+2ca\] \[↔(a-b)^2+(b-c)^2+(a-c)^2\ge 0 \quad (\quad luon\quad dung)\]
Dấu $"="$ xảy ra $⇔a=b=c=\sqrt{3}$