# 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:
$$\frac{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}{3} \ge \left( \frac{x}{y} \frac{y}{z} \frac{z}{x} \right) ^{\frac{1}{3}}=1.$$