Ukrainian National Mathematical Olympiad

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