Mathematical Reflections 2012, Issue 6 - Problem O249

Problem:
Find all triples $(x,y,z)$ of positive integers such that $$\dfrac{x}{y}+\dfrac{y}{z+1}+\dfrac{z}{x}=\dfrac{5}{2}.$$

Proposed by Titu Andreescu.

Solution:
By the AM-GM Inequality it must be $$3\sqrt[3]{\dfrac{z}{z+1}} \leq \dfrac{x}{y}+\dfrac{y}{z+1}+\dfrac{z}{x}=\dfrac{5}{2},$$ i.e. $\dfrac{z}{z+1} \leq \dfrac{125}{216},$ which gives $z=1$.
Using the AM-GM Inequality once again we obtain
$$2\sqrt{\dfrac{x}{2}} \leq \dfrac{x}{y}+\dfrac{y}{2}+\dfrac{1}{x}=\dfrac{5}{2},$$
i.e. $\dfrac{x}{2} \leq \dfrac{25}{16}$, which gives $x=1,2,3$.

(i) If $x=1$, then $\dfrac{1}{y}+\dfrac{y}{2}+1=\dfrac{5}{2}$ and solving for $y$ we get $y=1,2$.
(ii) If $x=2$, then $\dfrac{2}{y}+\dfrac{y}{2}+\dfrac{1}{2}=\dfrac{5}{2}$ and solving for $y$ we get $y=2$.
(iii) If $x=3$, then $\dfrac{3}{y}+\dfrac{y}{2}+\dfrac{1}{3}=\dfrac{5}{2}$, which gives no solution.

Therefore, all the triples of positive integers which satisfy the given equation are $$(1,1,1),(1,2,1),(2,2,1).$$