20th Turkish Mathematical Olympiad

The first term $\frac{x(2x-y)}{y(2z+x)}+1=\frac{2(x^2+yz)}{y(2z+x)}$. The similar transformation of two other terms yields: $$f(x,y,z)=\frac{x^2+yz}{y(2z+x)}+\frac{y^2+zx}{z(2x+y)}+\frac{z^2+xy}{x(2y+z)}\ge 2$$ By Cauchy-Schwarz inequality for positive $x_1,\dots ,x_n$ $$(x_1+\cdots +x_n)\left(\frac{a_1^2}{x_1}+\cdots +\frac{a_n^2}{x_n}\right)\ge (a_1+\cdots +a_n{)}^2(1)$$ Therefore, $$g(x,y,z)=\frac{x^2}{y(2z+x)}+\frac{y^2}{z(2x+y)}+\frac{z^2}{x(2y+z)}\ge \frac{(x+y+z{)}^2}{3(xy+yz+zx)}$$ $$h(x,y,z)=\frac{z}{2z+x}+\frac{x}{2x+y}+\frac{y}{2y+z}\ge \frac{(x+y+z{)}^2}{2(x^2+y^2+z^2)+xy+yz+zx}$$ Therefore, $$\begin{array}{rl}f& =g+h\ge (x+y+z{)}^2\left(\frac{1}{3(xy+yz+zx)}+\frac{1}{2(x^2+y^2+z^2)+xy+yz+zx}\right)\\ & =\frac{2(x+y+z{)}^4}{3(xy+yz+zx)(2(x^2+y^2+z^2)+xy+yz+zx)}\\ & =\frac{2(x+y+z{)}^4}{3(xy+yz+zx)(2(x+y+z{)}^2-3(xy+yz+zx))}\\ & \ge \frac{2(x+y+z{)}^4}{{\left(\frac{3(xy+yz+zx)+2(x+y+z{)}^2-3(xy+yz+zx)}{2}\right)}^2}=2\end{array}$$ (in the last inequality we applied AM-GM inequality). Done.