$\sum \frac{x^2y}{x+2y}=\sum \frac{x-\frac{x^2y}{x+2y}}{2}$, so we need to prove $\sum 4x^2\leq (x+y+z)^2+\sum\frac{4x^3}{x+2y}$ By AM-GM $\sum 4x^2<\sum (x(x+2y)+\frac{4x^3}{x+2y})$

Note that \[(x+y+z)^2 - \sum \frac{8x^2y}{x+2y} = \sum x(x+2y) - \sum \frac{8x^2y}{x+2y} = \sum \frac{x\left (x^2-4xy+4y^2 \right )}{x+2y} = \sum \frac{x(x-2y)^2}{x+2y}\ge 0\]Equality occur when $x=y=z=0$. Since $x,y,z>0$, we have \[(x+y+z)^2 - \sum\frac{8x^2y}{x+2y}>0 \Longrightarrow \sum \frac{x^2y}{x+2y} < \frac{(x+y+z)^2}{8}.\]