Question

Find the largest of three distinct positive integers with the least possible sum such that the sum of the reciprocals of any two integers among them is an integral multiple of the remaining one.

Answer 1

Let $x,y,z$ be three distinct positive integers satisfying the given conditions.
We may assume that $x<y<z$.
Thus we have three relations:
$\frac{1}{y}+\frac{1}{z}=\frac{a}{x},\frac{1}{z}+\frac{1}{x}=\frac{b}{y}=\frac{c}{z}$
for some positive integers $a,b,c$.
Thus $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{a+1}{x}=\frac{b+1}{y}=\frac{c+1}{z}=r$,
say.
Since $x<y<z$, we observe that $a<b<c$.
We also get $\frac{1}{x}=\frac{r}{a+1}$, $\frac{1}{y}=\frac{r}{b+1}$, $\frac{1}{z}=\frac{r}{c+1}$
Adding these, we obtain $r=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{r}{a+1}+\frac{r}{b+1}+\frac{r}{c=1}$,
or $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\mathrm{1.............}\left(1\right)$
Using $a<b<c$, we get
$1=\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}<\frac{3}{a+1}$
Thus $a<2$.
We conclude that $a=1$.
Putting this in the relation $\left(1\right)$, we get $\frac{1}{b+1}+\frac{1}{c+1}=1-\frac{1}{2}=\frac{1}{2}$
Hence $b<c$ gives $\frac{1}{2}<\frac{2}{b+1}$
Thus $b+1<4$ or $b<3$. since $b>a=1$,we must have $b=2$.
This gives $\frac{1}{c+1}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$,
or $c=5$.
Thus $x:y:z=a+1;b+1;c+1=2:3:6$.
Thus the required numbers with the least sum are $2,3,6$