Giải thích các bước giải:

Ta có:

$\dfrac{a^3}{(b+1)(c+1)}+\dfrac{b+1}8+\dfrac{c+1}8\ge 3\sqrt[3]{\dfrac{a^3}{(b+1)(c+1)}\cdot\dfrac{b+1}8\cdot\dfrac{c+1}8}$ 

$\to \dfrac{a^3}{(b+1)(c+1)}+\dfrac{b+1}8+\dfrac{c+1}8\ge \dfrac34a$

Tương tự:

$\dfrac{b^3}{(c+1)(a+1)}+\dfrac{c+1}8+\dfrac{a+1}8\ge\dfrac34b$

$\dfrac{c^3}{(a+1)(b+1)}+\dfrac{a+1}8+\dfrac{b+1}8\ge\dfrac34c$

Cộng vế với vế

$\to \dfrac{a^3}{(b+1)(c+1)}+\dfrac{b^3}{(c+1)(a+1)}+\dfrac{c^3}{(a+1)(b+1)}+\dfrac14(a+b+c)+\dfrac34\ge \dfrac34(a+b+c)$

$\to \dfrac{a^3}{(b+1)(c+1)}+\dfrac{b^3}{(c+1)(a+1)}+\dfrac{c^3}{(a+1)(b+1)}\ge \dfrac12(a+b+c)-\dfrac34$

$\to \dfrac{a^3}{(b+1)(c+1)}+\dfrac{b^3}{(c+1)(a+1)}+\dfrac{c^3}{(a+1)(b+1)}\ge \dfrac34$

Dấu = xảy ra khi $a=b=c=1$