Giải thích các bước giải:
Ta có :
$\dfrac{a^6}{b^2}+a^2b^2\ge 2\sqrt{\dfrac{a^6}{b^2}.a^2b^2}=2a^4$
$\dfrac{b^6}{a^2}+a^2b^2\ge 2\sqrt{\dfrac{b^6}{a^2}.a^2b^2}=2b^4$
$\to \dfrac{a^6}{b^2}+a^2b^2+\dfrac{b^6}{a^2}+a^2b^2\ge 2(a^4+b^4)$
$\to \dfrac{a^6}{b^2}+\dfrac{b^6}{a^2}+2a^2b^2\ge 2(a^4+b^4)$
$\to \dfrac{a^6}{b^2}+\dfrac{b^6}{a^2}\ge a^4+b^4+(a^4-2a^2b^2+b^4)$
$\to \dfrac{a^6}{b^2}+\dfrac{b^6}{a^2}\ge a^4+b^4+(a^2-b^2)^2\ge a^4+b^4$
