Giải thích các bước giải:
Ta có:
$A=(a+\dfrac1a)^2+(b+\dfrac1b)^2+(c+\dfrac1c)^2$
$\to A\ge \dfrac13(a+\dfrac1a+b+\dfrac1b+c+\dfrac1c)^2$
$\to A\ge \dfrac13(a+b+c+\dfrac1a+\dfrac1b+\dfrac1c)^2$
$\to A\ge \dfrac13(a+b+c+\dfrac9{a+b+c})^2$
$\to A\ge \dfrac13\cdot (1+\dfrac91)^2$
$\to A\ge \dfrac{100}{3}$
Dấu = xảy ra khi $a=b=c=\dfrac13$