# Proving $\frac{x}{y} +\frac{y}{z} + \frac{z}{x} \ge 3$ for positive $x,y,z$

Suppse $x,y,z$ 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?

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AM-GM inequality en.wikipedia.org/wiki/… – Ma Ming May 4 '13 at 23:51

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.

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I'd fix the ge to a \ge in the displayed inequality, but it's a nice solution. +1 – coffeemath May 4 '13 at 23:59
I feel so dumb... – cowchee May 5 '13 at 0:05