[tex]\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x} \geq 3[/tex]
Inegalitatea mediilor:
[tex]\left.\begin{cases}\dfrac{x}{y}+\dfrac{y}{z}\geq 2\sqrt{\dfrac{x}{y}\cdot \dfrac{y}{z}}\\ \dfrac{z}{x}+1 \geq 2\sqrt{\dfrac{z}{x}\cdot 1}\end{cases}\right|\Rightarrow \dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}+1 \geq 2\sqrt{\dfrac{x}{z}}+2\sqrt{\dfrac{z}{x}}[/tex]
[tex]\Rightarrow \dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}+1 \geq 2\left(\dfrac{\sqrt{x}}{\sqrt{z}}+\dfrac{\sqrt{z}}{\sqrt{x}}\right)[/tex]
[tex]\dfrac{\sqrt{x}}{\sqrt{z}}+\dfrac{\sqrt{z}}{\sqrt{x}}\geq 2\sqrt{\dfrac{\sqrt{x}}{\sqrt{z}}\cdot \dfrac{\sqrt{z}}{\sqrt{x}}} = 2\sqrt{1} = 2[/tex]
[tex]\Rightarrow \dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}+1 \geq 2\cdot 2[/tex]
[tex]\Rightarrow \boxed{\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x} \geq 3}[/tex]
