Proving $\frac{x}{y} +\frac{y}{z} + \frac{z}{x} \ge 3$ for positive $x,y,z$
Solution 1:
Divide through by 3. From AM-GM, we have for any positive $a,b,c$
$$\frac{a+b+c}{3} \ge (abc)^{\frac{1}{3}}$$
With $a=\frac{x}{y}, b=...$, this becomes:
$$\frac{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}{3} \ge \left( \frac{x}{y} \frac{y}{z} \frac{z}{x} \right) ^{\frac{1}{3}}=1.$$
Now multiply by 3 again to get the answer.