cannot be the value of the expression.

Question

Which of the following can not be the value of $x/y+y/z+z/x$. Where $x$, $y$, and $z$ are positive integers?

a) $4 $

b) $7/2$

c) $3$

d) $5/2$

Should I go through the options?

Answer 1

To begin, by the AM-GM inequality,

$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge3\sqrt{\frac{xyz}{yzx}}=3.$$

Since $\dfrac{5}{2}<3$, this cannot be the value of the sum.

Since we are only looking at the set of positive integers, we also need to give explicit constructions to exclude the other options (for instance, if one of your other options were $\pi$ we would never be able to get to it even though $\pi>3$).

Noting that $3=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}$ and $\frac{7}{2}=\frac{1}{2}+\frac{2}{1}+\frac{1}{1}$, we eliminate those choices.

The option $4$, however, is interesting. I find that I am neither able to give an explicit construction nor able to prove that it is not achievable.

EDIT: After much hullabaloo, it has been discovered that there are also no integer solutions to $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=4$. This comes from A.V. Bondarenko. Investigation of a class of Diophantine equations. (Russian. English, Ukrainian summary) Ukraïn. Mat. Zh. 52 (2000), no. 6, 831--836;

For more information, please see this question.

Answer 2

Let $x \le y \le z$, wlog.

By the rearrangement inequality, $x/y + y/z + z/x \ge x/x + y/y + z/z = 3$.

So 5/2, which is less than 3, is not possible.