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?
 A: 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):
$$x\frac{1}{y}+z\frac{1}{x}+y\frac{1}{z}\ge x\frac{1}{x}+z\frac{1}{z}+y\frac{1}{y}=3$$
A: One can use rearrangement inequality since $(x,y,z)$ is certainly in pairwise opposite order as compared to $(\frac{1}{x},\frac{1}{y},\frac{1}{z})$ for positive $x,y,z$. In turn, rearrangement inequality follows from repeatedly applying $ab+cd \ge ad+cb$ whenever $a \ge c$ and $b \ge d$.
By the way your symmetric inequality is actually too weak to be able to derive your original desired inequality. In general cyclic inequalities need far stronger tools than symmetric ones. (In this case of course just one AM-GM would have done it.)
A: Hint. 
Consider your symmetric quantity as a function of $z$ and look at its minimum for $z>0$
