30th Junior Turkish Mathematical Olympiad

By applying AM-GM inequality we get $$xy+y+2z=xy+y+z+z\ge 4\sqrt[4]{x{y}^2{z}^2}$$ Since $0<x\le 1$ we have $x\ge x^2$ and hence $$4\sqrt[4]{x{y}^2{z}^2}\ge 4\sqrt[4]{x^2{y}^2{z}^2}=4\sqrt{xyz}$$ and we are done.

---

Alternative solution.

Since $0<x\le 1$ we have $y\ge xy$ and hence $$xy+y+2z\ge 2xy+2z\ge 4\sqrt{xyz}$$ by AM-GM inequality.