Prove that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \geq 3$ for $x,y,z>0$ By considering that $$\frac{x}{y}+\frac{y}{x} \geq 2$$I can show that $$\frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y} \geq 6$$
But how would one go from here to prove the required result? It feels like I'm almost there but I can't quite see how to finish it.
I can see that it can be killed off almost instantly using AM/GM of degree 3, but how would one acquire this result without the use of such a tool?

By AM-GM inequality, since $x,y,z>0$:
$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \geq 3\sqrt[3]{\frac{x}{y}\frac{y}{z}\frac{z}{x}}=3$$
with equality iff $\frac{x}{y}=\frac{y}{z}=\frac{z}{x}$, i.e. $x=y=z$.
Alternatively, rearrangement inequality. $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ is cyclic - two cases:

*

*wlog $x\ge y\ge z$. Then $\frac{1}{z}\ge\frac{1}{y}\ge \frac{1}{x}$, so (RHS oppositely sorted):
$$x\frac{1}{y}+y\frac{1}{z}+z\frac{1}{x}\ge x\frac{1}{x}+y\frac{1}{y}+z\frac{1}{z}=3$$


*wlog $x\ge z\ge y$. Then $\frac{1}{y}\ge\frac{1}{z}\ge\frac{1}{x}$, so (RHS oppositely sorted):
$$x\frac{1}{y}+z\frac{1}{x}+y\frac{1}{z}\ge x\frac{1}{x}+z\frac{1}{z}+y\frac{1}{y}=3$$
