Giải thích các bước giải:
Ta có:
$\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge 2\sqrt{ab}\cdot 2\sqrt{bc}\cdot 2\sqrt{ca}=8abc$
Mà $4=a+b+c\ge 3\sqrt[3]{abc}$
$\to 64\ge 27abc$
$\to abc\le \dfrac{64}{27}$
$\to a^2b^2c^2\le \left(\dfrac{64}{27}\right)^2<8$
$\to 8abc\ge a^3b^3c^3$
$\to \left(a+b\right)\left(b+c\right)\left(c+a\right)>a^3b^3c^3$