# Prove that the inequality is true

Prove that forl all positive real numbers $$x, y, z$$ we have that $$(x^3 + y^3 + z^3)^2 \geq 3(x^2y^4 + y^2z^4 + z^2x^4)$$. I tried to apply Cebasev, Muirhead but doesn't work.

$$\begin{eqnarray*} y^3(2x+y)(x-y)^2+z^3(2y+z)(y-z)^2+x^3(2z+x)(z-x)^2 \geq 0 \end{eqnarray*}$$ Now rearrange.
Because $$(x^3+y^3+z^3)^2-3(x^4z^2+y^4x^2+z^4y^2)=$$ $$=\sum_{cyc}(x^6+2x^3y^3-3x^2y^4)=\sum_{cyc}x^2(x^4+2xy^3-3y^4)=$$ $$=\sum_{cyc}x^2(x^4-x^3y+x^3y-x^2y^2+x^2y^2-xy^3+3xy^3-3y^4)=$$ $$=\sum_{cyc}x^2(x-y)(x^3+x^2y+xy^2+3y^3)=$$ $$=\sum_{cyc}\left((x-y)(x^5+x^4y+x^3y^2+3x^2y^3-(x^6-y^6)\right)=$$ $$=\sum_{cyc}(x-y)(2x^2y^3-xy^4-y^5)=\sum_{cyc}y^3(x-y)^2(2x+y)\geq0.$$ Also, there is a proof by Rearrangement and AM-GM, but it's much more complicated.