Giải thích các bước giải:
Ta có:
$3(a^3+b^3)=(a^3+b^3+b^3)+(a^3+a^3+b^3)\ge 3\sqrt[3]{a^3\cdot b^3\cdot b^3}+3\sqrt[3]{a^3\cdot a^3\cdot b^3}$
$\to 3(a^3+b^3)\ge 3ab^2+3ab^2$
$\to a^3+b^3\ge ab(a+b)$
$\to 2a^3+b^3+6=(a^3+b^3)+(a^3+1+1)+4$
$\to 2a^3+b^3+6\ge ab(a+b)+3a+4$
$\to 2a^3+b^3+6\ge ab(a+b)+1+3a+3$
$\to 2a^3+b^3+6\ge ab(a+b)+abc+3a+3$ vì $abc=1$
$\to 2a^3+b^3+6\ge ab(a+b+c)+3a+3$
Lại có: $a+b+c\ge 3\sqrt[3]{abc}=3$
$\to 2a^3+b^3+6\ge 3ab+3a+3$
$\to \dfrac{1}{2a^3+b^3+6}\le \dfrac{1}{3ab+3a+3}$
$\to \dfrac{1}{2a^3+b^3+6}\le \dfrac{1}{3(ab+a+1)}$
Tương tự ta chứng minh được:
$\dfrac{1}{2b^3+c^3+6}\le\dfrac{1}{3(bc+b+1)}$
$\dfrac{1}{2c^3+a^3+6}\le\dfrac{1}{3(ca+c+1)}$
$\to \dfrac{1}{2a^3+b^3+6}+\dfrac{1}{2b^3+c^3+6}+\dfrac{1}{2c^3+a^3+6}\le\dfrac{1}{3(ab+a+1)}+\dfrac{1}{3(bc+b+1)}+\dfrac{1}{3(ca+c+1)}$
$\to \dfrac{1}{2a^3+b^3+6}+\dfrac{1}{2b^3+c^3+6}+\dfrac{1}{2c^3+a^3+6}\le\dfrac13(\dfrac{1}{ab+a+1}+\dfrac{1}{bc+b+1}+\dfrac{1}{ca+c+1})$
Mà $A=\dfrac{1}{ab+a+1}+\dfrac{1}{bc+b+1}+\dfrac{1}{ca+c+1}$
$\to A=\dfrac{1}{ab+a+1}+\dfrac{a}{abc+ab+a}+\dfrac{ab}{abca+abc+ab}$
$\to A=\dfrac{1}{ab+a+1}+\dfrac{a}{1+ab+a}+\dfrac{ab}{a+1+ab}$ vì $abc=1$
$\to A=1$
$ \to \dfrac{1}{2a^3+b^3+6}+\dfrac{1}{2b^3+c^3+6}+\dfrac{1}{2c^3+a^3+6}\le\dfrac13$
Dấu = xảy ra khi $a=b=c=1$
