Giải thích các bước giải:
Câu 1.$a+b=(\dfrac{a}{3}+\dfrac{a}{3}+\dfrac{b}{3})+(\dfrac{b}{3}+\dfrac{b}{3}+\dfrac{a}{3})\ge 3\sqrt[3]{\dfrac{a}{3}.\dfrac{a}{3}.\dfrac{b}{3}}+3\sqrt[3]{\dfrac{b}{3}.\dfrac{b}{3}.\dfrac{a}{3}}\\
\rightarrow a+b\ge \sqrt[3]{a^2b}+\sqrt[3]{ab^2}\quad\forall a,b>0$
Câu 2.
$(a+b+c)(ab+bc+ca-2)\ge 3\sqrt[3]{abc}.(3\sqrt[3]{ab.bc.ca}-2)=3\\
\rightarrow 2+2(a+b+c) \le (a+b+c)(ab+bc+ca)-abc\\
\rightarrow 2+2(a+b+c) \le (a+b)(b+c)(c+a)\\
\rightarrow (a+b)(b+c)+a+b+b+c+1 + (b+c)(c+a)+b+c+c+a+1 + (c+a)(a+b)+c+a+a+b+1
≤ (a+b)(b+c)(c+a) + (a+b)(b+c) + (b+c)(c+a) + (c+a)(a+b) +a+b+b+c+c+a+1 \\
\rightarrow (a+b+1)(b+c+1) + (b+c+1)(c+a+1) + (c+a+1)(a+b+1) ≤ (a+b+1)(b+c+1)(c+a+1) \\
\rightarrow \dfrac{1}{a+b+1}+\dfrac{1}{b+c+1}+\dfrac{1}{c+a+1}\le 1$
