Solution:
Notice that $\sum_{cyc} \frac{x+2y}{z+2x+3y} = \sum_{cyc} \left(1 - \frac{x+y+z}{z+2x+3y}\right) = 3 - (x+y+z) \sum_{cyc} \frac{1}{z+2x+3y}$.

We have to prove that $3 - (x+y+z) \sum_{cyc} \frac{1}{z+2x+3y} \leq \frac{3}{2}$ or $\frac{3}{2(x+y+z)} \leq \sum_{cyc} \frac{1}{z+2x+3y}$.

By Cauchy-Schwarz we obtain $\sum_{cyc} \frac{1}{z+2x+3y} \geq \frac{(1+1+1)^2}{\sum_{cyc}(z+2x+3y)} = \frac{3}{2(x+y+z)}$.

Because the inequality is homogeneous, we can take $x+y+z=1$.

Denote $x+2y = a$, $y+2z = b$, $z+2x = c$. Hence, $a+b+c = 3(x+y+z) = 3$.

We have $(k-1)^2 \geq 0 \Leftrightarrow (k+1)^2 \geq 4k \Leftrightarrow \frac{k+1}{4} \geq \frac{k}{k+1}$ for all $k > 0$.

Hence $\sum_{cyc} \frac{x+2y}{z+2x+3y} = \sum \frac{a}{1+a} \leq \sum \frac{a+1}{4} = \frac{a+b+c+3}{4} = \frac{3}{2}$.