Giải thích các bước giải:
Ta có :
$(a+\dfrac{1}{a})^2+(b+\dfrac{1}{b})^2+(c+\dfrac{1}{c})^2$
$=(a^2+b^2+c^2)+(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2})+6$
$\ge \dfrac{1}{3}(a+b+c)^2+\dfrac{1}{3}(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^2+6$
$\ge \dfrac{1}{3}(a+b+c)^2+\dfrac{1}{3}(\dfrac{9}{a+b+c})^2+6$
$=\dfrac{100}{3}$
Dấu = xảy ra khi $a=b=c=\dfrac{1}{3}$