Proving that $\frac {x}{ y} + \frac {y}{z} + \frac {z}{x} = 2$ has no natural solutions

Question

I'm having trouble with this problem:

Prove, that the equation $ \frac {x}{ y} + \frac {y}{z} + \frac {z}{x} = 2$ has no solutions that are natural numbers.

The only thing I've managed to come up with is that no two number can be equal in this equation (if all three would be equal the sum would obviously be 3 and if two of the numbers were equal and the other one either smaller or bigger, the sum would be more than two, because one of the fractions would be 1, one smaller and one bigger than 1), but I don't know how to prove that no natural solutions exist, so I'd appreciate any help with this.

Answer 1

Use AM–GM inequality for three numbers: $\dfrac{x}{y}$, $\dfrac{y}{z}$, $\dfrac{z}{x}$.

Answer 2

Let $c$ be some positive number and let's consider the equation $a + 1/a = c.$ When does this equation have a solution in the real numbers? Exactly when the polynomial $a^2 - ca + 1$ has a real root or equivalently when $c^2 - 4 \geq 0.$ So in order for $a + 1/a = c$ to have a solution $c$ must be greater than or equal to $2.$ It follows if $x,y,z$ is a solution to your equation and $x \leq y \leq z,$

$$\frac{y}{z} = 2 - \left(\frac xy + \frac zx\right) \leq 2 - \left(\frac xz + \frac zx\right) \leq 0,$$

a contradiction.