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Prove that the following equation :

$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=2$

$x,y,z\in\mathbb{N}$

Has no solution in $\mathbb{N^{3}}$

I'm going to use the hint are given :

$\left(\frac{a+b+c}{3}\right)^{3}\geq abc$

But how I use it ??

For example if we take :

$\frac{x}{y}=\frac{y}{z}=\frac{z}{x}$ then $\prod \frac{x}{y}=1$

this mean : $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq 3$

Bill Dubuque
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Dinize markob
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2 Answers2

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The hint you mention is the Inequality of arithmetic and geometric means, specifically applied to any $3$ non-negative real values.

With your case, as you stated, you have $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \ge 3$. Thus, $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 2$ can't be true. Note this is true in general for any $x,y,z \gt 0$ with $x,y,z \in \mathbb{R}$, not just those where $x,y,z \in \mathbb{N}$.

Update: As mentioned in this answer, I didn't notice originally, so I didn't address in my solution, that you used a specific case of $\frac{x}{y} = \frac{y}{z} = \frac{z}{x}$ in your question. Nonetheless, my final sentence does make clear the result of $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \ge 3$ is actually true for all positive, real numbers, not just positive integers.

John Omielan
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You used the inequality in the special case of $\frac{x}{y} = \frac{y}{z} = \frac{z}{x}$. You need to prove the result in general without this assumption, as follows:- $$\left(\frac{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}{3}\right)^3 \ge \frac{x}{y}\frac{y}{z}\frac{z}{x}=1.$$ So $$\frac{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}{3} \ge 1.$$

Then as per the solution of @John Omielan.