We will prove reverse i.e. we will assume $4x+y<4z$ and prove $2(x^3+y^3+z^3)+15(xy^2+yz^2+zx^2)<16(x^2y+y^2z+z^2x)+2xyz$ ($\bigstar$). Let $z=2x+b, y=4x+c.$ $y\geq 2z\geq 4x\Rightarrow c\geq 2b\geq 0$ $(1)$ $4x+y<4z\Rightarrow c<4b$ $(2)$ Mix $(1)$ and $(2)$ $4b>c\geq 2b>0$ $4>k=\frac{c}{b} \geq 2$ Case 1. $k>2$ Replace $(x,y,z)$ with $(x,4x+c,2x+b)$ in ($\bigstar$) and rearrange it as a monovariable polynomial for $x$. Now we want to prove below $49bx^2-(56b^2+7c^2-70bc)x-(2b^3+2c^3+15b^2c-16bc^2)>0$ ($\bigstar \bigstar$) If we prove $\Delta<0$, with using $49b>0$, it end. $\frac{\Delta}{49}=c^4-12c^3b+52c^2b^2-100cb^3+72b^4<0$ $k^4-12k^3+52k^2-100k+72=P(k)<0$ $P(k)=(k-2)(k^3-10k^2+32k-36)=(k-2)Q(k)<0$ $Q'(k)=3k^2-20k+32=(3k-8)(k-4)$ $Q(k)$ is increasing on interval $(2,\frac{8}{3})$ and decreasing on interval $(\frac{8}{3},4)$ so $Q(k)\leq Q(\frac{8}{3})<0$. Done. Case 2. $k=2$ This case equal to $c=2b$ in ($\bigstar \bigstar$). It become: $49bx^2+56b^2x+16b^3=b(7x+4b)^2>0$ If $x\neq \frac{-4b}{7}$ it end. If $x=\frac{-4b}{7}$; $y=4x+2b=\frac{-2b}{7}$ and $z=2x+b=\frac{-b}{7}$. Equality is homogenous so $(x,y,z)=(-4t,-2t,-t), t>0$. [I think there is small mistake about problem. Because for $(x,y,z)=(-4t,-2t,-t)$; $y=-2t\geq 2z=-2t\geq 4x=-16t$ and $-656t^3\geq -656t^3$ are true but $4x+y=-18t\geq 4z=-4t$ is not true. But this is the only thing that do not satisfy.]

This problem has been edited: [Corrected]: Let $x$, $y$, $z$ be real numbers satisfying $y>2z>4x$ and $$ 2(x^3+y^3+z^3) + 15(xy^2+yz^2+zx^2) > 16(x^2y+y^2z+z^2x) + 2xyz.$$Show that $4x + y > 4z.$