# 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:

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

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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 $a=b=c.$)

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Just by sum of square and Schur's inequality: $$3(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)-(ab+bc+ca)^3\\ = \sum_{cyc} (a^{3}b+a^{3}c+a^{2}bc)(b-c)^2+3abc\sum_{cyc}a(a-b)(a-c) \geq 0$$

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