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:
$$(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)>\frac{(ab+bc+ca)^3}{3}$$

By Cauchy-Schwarz:
$(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2) \ge (a^{1.5}b^{1.5}+b^{1.5}c^{1.5}+c^{1.5}a^{1.5})^2$
By the power mean inequality:
$(a^{1.5}b^{1.5}+b^{1.5}c^{1.5}+c^{1.5}a^{1.5})^2 \ge \frac{(ab+bc+ca)^3}{3}.$
(Note: The > sign in your post is false; let $a=b=c.$)
