jmc

Let $a=2x,$ $b=y,$ and $c=2z.$ Then $x=\frac{a}{2},$ $y=b,$ and $z=\frac{c}{2},$ so $$\begin{array}{rl}\frac{4z}{2x+y}+\frac{4x}{y+2z}+\frac{y}{x+z}& =\frac{2c}{a+b}+\frac{2a}{b+c}+\frac{b}{\frac{a}{2}+\frac{c}{2}}\\ & =\frac{2c}{a+b}+\frac{2a}{b+c}+\frac{2b}{a+c}\\ & =2\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right).\end{array}$$Let $$S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}.$$Then $$\begin{array}{rl}S+3& =\frac{a}{b+c}+1+\frac{b}{a+c}+1+\frac{c}{a+b}+1\\ & =\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\\ & =(a+b+c)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\\ & =\frac{1}{2}(2a+2b+2c)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\\ & =\frac{1}{2}[(b+c)+(a+c)+(a+b)]\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right).\end{array}$$By Cauchy-Schwarz, $$[(b+c)+(a+c)+(a+b)]\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\ge (1+1+1{)}^2=9,$$so $$S\ge \frac{9}{2}-3=\frac{3}{2},$$and $$\frac{4z}{2x+y}+\frac{4x}{y+2z}+\frac{y}{x+z}\ge 2S=3.$$Equality occurs when $a=b=c,$ or $2x=y=2z,$ so the minimum value is $\boxed{3}.$