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.
 A: In case you want to expand your list of basic inequalities, by Holder's inequality
$$LHS=(1+1+1)(x^2y+y^2z+z^2x)(zx^2+xy^2+yz^2) \ge \left(\sqrt[3]{x^4yz}+\sqrt[3]{xy^4z}+\sqrt[3]{xyz^4}\right)^3=RHS$$
A: The inequality is equivalent to 
$$3\sum x^3y^3 +2\sum xy^4z+3x^2y^2z^2\ge 3\sum(xy^3z^2+x^2y^3z) =3\sum xy^3z(x+z)$$
AM-GM gives
$$x^3y^3+xy^4z+x^2y^2z^2\ge 3 x^2y^3z$$
$$y^3z^3+xy^4z+x^2y^2z^2\ge 3xy^3z^2$$
Summing up we have
$$2\sum x^3y^3+2\sum xy^4z+6x^2y^2z^2\ge 3\sum (x^2y^3z+xy^3z^2),$$
or 
$$\sum x^3y^3+\sum xy^4z+3x^2y^2z^2\ge \frac32\sum(x^2y^3z+xy^3z^2).\tag{1}$$
Now, note that 
$$x^3+y^3\ge xy^2+yx^2.\tag{2}$$
Multiplying (2) with $xyz$ gives
$$\sum xy^4z\ge \frac12\sum(x^2y^3z+x^3y^2z)\tag{3}.$$
Multiplying (2) with $z^3$ gives
$$\sum (x^3z^3+y^3z^3)\ge \sum(xy^2z^3+x^2yz^3).\tag{4}$$
The inequality follows from (1), (3), and (4).
Note: It may not be a good book to work with after all. And I'm sure there are other non brute-force proofs.
A: Since $\let\geq\geqslant\let\leq\leqslant$the inequality is homogenous, we may assume $x+y+z=1$. The inequality can be rewritten as
$$3\cdot\sum_{cyc}\frac xy\sum_{cyc}\frac yx\geq\frac1{xyz}.$$
By weighted AM-GM,
$$\sum_{cyc}\frac xy\geq(x+y+z)\sqrt[x+y+z]{\frac1{y^xx^zz^y}}.$$
Combining a similar inequality for the other factor yields 
$$LHS\geq3\frac{x^xy^yz^z}{xyz},$$
so it suffices to prove $3x^xy^yz^z\geq1$, which is true by Jensen's inequality in the form
$$x\log x+y\log y+z\log z\geq3\frac{x+y+z}3\log\frac{x+y+z}3=-\log 3$$
because $x\log x$ is convex on the interval $]0,1]$.
