Prove $4(x + y + z)^3 > 27(x^2y + y^2z + z^2x)$ Prove that, for all positive real numbers $x$, $y$ and $z$,
$$4(x + y + z)^3 > 27(x^2y + y^2z + z^2x)$$
I've tried expressing it as a sum of squares, but haven't got anywhere. 
Hints are also welcome.

Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.
Thus, by Rearrangement and AM-GM we obtain:
$$x^2y+y^2z+z^2x=x\cdot xy+y\cdot yz+z\cdot zx\leq a\cdot ab+b\cdot ac+c\cdot bc=$$
$$=a^2b+abc+c^2b\leq a^2b+2abc+c^2b=b(a+c)^2=4\cdot b\left(\frac{a+c}{2}\right)^2\leq$$
$$\leq4\left(\frac{b+2\cdot\frac{a+c}{2}}{3}\right)^3=\frac{4}{27}(a+b+c)^3=\frac{4}{27}(x+y+z)^3.$$
