prove this inequality with $x^3+y^3+z^3$ 
Let $x,y,z>0$. Show that
  $$x^3+y^3+z^3+2(xy^2+yz^2+zx^2)\ge 3(x^2y+y^2z+z^2x).$$

I have a solution by using $y=x+a,z=x+a+b$; my question is: can this be solved using simple methods (such as AM-GM, C-S, and so on)?
 A: We need to prove that $$\sum_{cyc}(x^3-3x^2y+2x^2z)\geq0$$ or
$$\sum_{cyc}(2x^3-6x^2y+4x^2z)\geq0$$ or 
$$\sum_{cyc}(2x^3-x^2y-x^2z)\geq5\sum_{cyc}(x^2y-x^2z)$$ or
$$\sum_{cyc}(x+y)(x-y)^2\geq5(x-y)(x-z)(y-z).$$
Since our inequality is cyclic, we can assume that $x=\max\{x,y,z\}$ and since for $x\geq z\geq y$ we have $$(x-y)(x-z)(y-z)\leq0,$$ it's enough to prove our inequality for $x\geq y\geq z$.
Now, easy to see that if we'll change $(x,y,z)$ on $(x-h,y-h,z-h)$ for all $0\leq h<z$ 
so $(x-y)^2,$ $(x-z)^2$, $(y-z)^2$ and $(x-y)(x-z)(y-z)$ are not changed and the left side of the inequality 
$$\sum_{cyc}(x+y)(x-y)^2\geq5(x-y)(x-z)(y-z)$$ decreases.
Thus, it's enough to prove our inequality for $h\rightarrow z^-$, 
which says that it's enough to prove our inequality for $z\rightarrow0^+$.
Id est, we need to prove that
$$x^3-3x^2y+2xy^2+y^3\geq0,$$ which we can prove by AM-GM:
$$x^3-3x^2y+2xy^2+y^3=\frac{1}{2}(8\cdot\frac{x^3}{4}+6\cdot\frac{2xy^2}{3}+2y^3)-3x^2y\geq$$
$$\geq\frac{1}{2}\cdot15\sqrt[15]{\left(\frac{x^3}{4}\right)^8\left(\frac{2xy^2}{3}\right)^6\cdot2y^3}-3x^2y=\left(\frac{15}{2\sqrt[5]{72}}-3\right)x^2y>0.$$
