$4b)a^3+2\geq 3a,b^3+2\geq 3b,c^3+2\geq 3c\Rightarrow a^3+b^3+c^3\geq a+b+c+2(a+b+c)-6;a+b+c\geq 3\sqrt[3]{abc}=3\Rightarrow \Leftrightarrow a^3+b^3+c^3\geq a+b+c$
$4c)2a^3+2b^3+2c^3+3=(2a^3+1)+(2b^3+1)+(2c^3+1)\geq 3a^2+3b^2+3c^2;a^2+b^2+c^2\geqslant 3\sqrt[3]{a^2b^2c^2}=3\Rightarrow 2a^3+2b^3+2c^3\geq 2(a^2+b^2+c^2)\Rightarrow a^3+b^3+c^3\geq a^2+b^2+c^2$
- MrDat yêu thích