Prove $3(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\ge xyz(x+y+z)^3$ 
if $x,y,z$ are positive real numbers,Prove:$$3(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\ge xyz(x+y+z)^3$$
Additional info:$\sum_{cyc}$ denotes sums over cyclic permutations of the symbols $x,y,z$. I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.

Things I have done: Writing Left hand side in expanded form:$$LHS=3(\sum_{cyc}x^3y^3+\sum_{cyc}xy^4z+3x^2y^2z^2)$$
Right now I don't have good idea were to start(Is expanding and looking for AM-GM is good?).So any hint is appreciated.
Source : This inequality can be found as Problem 42, page 13, in the book "Old and New inequalities" by Titu Andreescu, where it is attributed to Manlio Marangelli.

In case you want to expand your list of basic inequalities, by Holder's inequality
$$LHS=(1+1+1)(x^2y+y^2z+z^2x)(zx^2+xy^2+yz^2) \ge \left(\sqrt[3]{x^4yz}+\sqrt[3]{xy^4z}+\sqrt[3]{xyz^4}\right)^3=RHS$$
