Giải thích các bước giải:
Ta có:
$\dfrac{a^3}{(b+2)^2}+\dfrac{b+2}{8}+\dfrac{b+2}{8}\ge 3\sqrt[3]{\dfrac{a^3}{(b+2)^2}.\dfrac{b+2}{8}.\dfrac{b+2}{8}}=\dfrac{3}{4}a$
$\rightarrow \dfrac{a^3}{(b+2)^2}\ge \dfrac{3}{4}a-\dfrac{b+2}{4}$
Tương tự ta chứng minh được
$ \dfrac{b^3}{(a+2)^2}\ge \dfrac{3}{4}b-\dfrac{a+2}{4}$
$\rightarrow \dfrac{a^3}{(b+2)^2}+\dfrac{b^3}{(a+2)^2}\ge \dfrac{3}{4}a-\dfrac{b+2}{4}+\dfrac{3}{4}b-\dfrac{a+2}{4}$
$\rightarrow \dfrac{a^3}{(b+2)^2}+\dfrac{b^3}{(a+2)^2}\ge \dfrac{1}{2}(a+b)-1=1$