58th Ukrainian National Mathematical Olympiad

We prove it by using Cauchy–Schwarz inequality: $$\left(\frac{y}{2x+y}+\frac{z}{2y+z}+\frac{x}{2z+x}\right)(x+y+z{)}^2=\left(\frac{y}{2x+y}+\frac{z}{2y+z}+\frac{x}{2z+x}\right)\left(y(2x+y)+z(2y+z)+x(2z+x)\right)\ge {\left(\sqrt{\frac{y}{2x+y}}\cdot \sqrt{y(2x+y)}+\sqrt{\frac{z}{2y+z}}\cdot \sqrt{z(2y+z)}+\sqrt{\frac{x}{2z+x}}\cdot \sqrt{x(2z+x)}\right)}^2=(x+y+z{)}^2.$$