Giải thích các bước giải:
Ta có:
$(a+\dfrac1a)^2+(b+\dfrac1b)^2+(c+\dfrac1c)^2$
$=(a^2+2+\dfrac{1}{a^2})+(b^2+2+\dfrac{1}{b^2})+(c^2+2+\dfrac{1}{c^2})$
$=(a^2+b^2+c^2)+(\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2})+6$
$\ge \dfrac{(a+b+c)^2}3+\dfrac13(\dfrac1a+\dfrac1b+\dfrac1c)^2+6$
$\ge \dfrac{(a+b+c)^2}3+\dfrac13(\dfrac9{a+b+c})^2+6$
$\ge \dfrac{(a+b+c)^2}3+\dfrac{27}{(a+b+c)^2}+6$
$\ge \dfrac{1}3+\dfrac{27}{1}+6$
$\ge 33+\dfrac13$
$>33$