Prove that, for all non-negative real numbers $x,y,z$ that satisfy $x+y+z=1, x^2y+y^2z+z^2x≤4/27$ Prove that, for all non-negative real numbers $x, y, z$ that satisfy $x + y + z = 1$,
$$x^2 y + y^2 z + z^2 x \leq \frac {4}{27}
$$
I'm having trouble with this question. I suspect it may have a fairly simple proof using the AM-GM inequality and certain substitutions, however, I have been unable to complete such a proof. 

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