58th Ukrainian National Mathematical Olympiad

Using this equation and the inequality between the arithmetic mean and the geometric mean, we have $$xy+yz+zx=xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{xyz{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}^2}{x+y+z}=\frac{xyz\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2x+2y+2z}{x+y+z}\ge \frac{xyz\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)}{x+y+z}+2=3$$