Ta có:
$\dfrac{a^4+b^4}{a^2+ab+b^2}=\dfrac{2(a^4+b^4)}{2(a^2+b^2)+2ab}\ge\dfrac{(a^2+b^2)^2}{2(a^2+b^2)+a^2+b^2}=\dfrac{a^2+b^2}{3}$
Tương tự: $\dfrac{b^4+c^4}{b^2+bc+c^2}\ge\dfrac{b^2+c^2}{3};\dfrac{c^4+a^4}{c^2+ca+a^2}\ge\dfrac{c^2+a^2}{3}$
Suy ra:
$\dfrac{a^4+b^4}{a^2+ab+b^2}+\dfrac{b^4+c^4}{b^2+bc+c^2}+\dfrac{c^4+a^4}{c^2+ca+a^2}\ge\dfrac{2}{3}(a^2+b^2+c^2)\ge 2\sqrt[3]{a^2b^2c^2}=2$
Dấu bằng xảy ra khi: $a=b=c=1$