<s>[INST] Proof of $(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)>\frac{(ab+bc+ca)^3}{3}$ For positive real numbers $a$, $b$ and $c$, how do we prove that:
$$(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)>\frac{(ab+bc+ca)^3}{3}$$
 [/INST] Just by sum of square and Schur's inequality:
$$
3(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)-(ab+bc+ca)^3\\
=
\sum_{cyc} (a^{3}b+a^{3}c+a^{2}bc)(b-c)^2+3abc\sum_{cyc}a(a-b)(a-c) \geq 0
$$
</s>