Giải thích các bước giải:
Ta có :
$(\dfrac{a^2}{b+2c}+\dfrac{b^2}{c+2a}+\dfrac{c^2}{a+2b})((b+2c)+(c+2a)+(a+2b))$
$\ge (\sqrt{\dfrac{a^2}{b+2c}.(b+2)}+\sqrt{\dfrac{b^2}{c+2a}.(c+2a)}+\sqrt{\dfrac{c^2}{a+2b}.(a+2b)})^2$
$=(a+b+c)^2$
$\to (\dfrac{a^2}{b+2c}+\dfrac{b^2}{c+2a}+\dfrac{c^2}{a+2b})(3(a+b+c))\ge (a+b+c)^2$
$\to \dfrac{a^2}{b+2c}+\dfrac{b^2}{c+2a}+\dfrac{c^2}{a+2b}\ge \dfrac{a+b+c}3$