Giải thích các bước giải:
a.Ta có :
$a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)((a+b)^2-3ab)\ge (a+b)((a+b)^2-\dfrac{3(a+b)^2}{4})=\dfrac{(a+b)^3}{4}$
$\to \dfrac{a^3+b^3}{2}\ge (\dfrac{a+b}{2})^3$
b.Ta có :
$\dfrac{a^6}{b^2}+\dfrac{b^6}{a^2}$
$=\dfrac{a^6}{b^2}+a^2b^2+\dfrac{b^6}{a^2}+a^2b^2-2a^2b^2$
$\ge 2\sqrt{\dfrac{a^6}{b^2}.a^2b^2}+2\sqrt{\dfrac{b^6}{a^2}.a^2b^2}-2a^2b^2$
$\ge 2a^4+2b^4-2a^2b^2$
$\ge a^4+b^4+(a^2-b^2)^2$
$\ge a^4+b^4$
