Proving $\frac{x}{y} +\frac{y}{z} + \frac{z}{x} \ge 3$ for positive $x,y,z$ 
Suppose $x,y,z$ are real positive numbers, prove that:
$$\dfrac{x}{y} +\dfrac{y}{z} + \dfrac{z}{x} \ge 3$$
with equality when $x=y=z$.

Can someone help me find an easier solution?
I started with assume only two are equal, without loss of generality, $x=y$. Then we get
$$1 + \dfrac{x}{z} + \dfrac{z}{x}$$
$$1 + \dfrac{x^2 + z^2}{xz} = 1+ \dfrac{(x-z)^2 + 2xz}{xz} = 1 + 2 + \dfrac{(x-z)^2}{xz} \ge 3$$
To prove the general result I considered using a derivative with respect to $y$ to show that, starting with $x=y$ and increasing $y$ by a tiny amount, will show the equation's derivative as positive. I haven't gotten this to work yet. Any ideas?

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.
