Giải thích các bước giải:
Ta có : $\dfrac{b^3+c^3}{2}\ge \left(\dfrac{b+c}{2}\right)^3\to b^3+c^3\ge \dfrac{\left(b+c\right)^3}{4}$
$\to 8=a^3+b^3+c^3\ge a^3+\dfrac{\left(b+c\right)^3}{4}\ge 2\sqrt{a^3.\dfrac{\left(b+c\right)^3}{4}}$
$\to \sqrt{a^3.\dfrac{\left(b+c\right)^3}{4}}\le 4$
$\to a^3.\dfrac{\left(b+c\right)^3}{4}\le 16$
$\to \dfrac{\left(a\left(b+c\right)\right)^3}{4}\le 16$
$\to \left(a\left(b+c\right)\right)^3\le 64$
$\to a\left(b+c\right)\le 4$
Dấu = xảy ra khi $b=c,a^3=\dfrac{\left(b+c\right)^3}{4}=2b^3\to b=c=\sqrt[3]{2},a=\sqrt[3]{4}$
