Indija mo 2011

We first observe that $(1,a,b)$ is not a solution whenever $1<a<b$. Otherwise we should have $\frac{1}{a}+\frac{1}{b}=l\cdot \frac{1}{1}=l$ for some integer $l$. Hence we obtain $\frac{a+b}{ab}=l$ showing that $a|b$ and $b|a$. But then $a=b$ contradicting $a\ne b$. Thus the least number should be $2$. It is easy to verify that $(2,3,4)$ and $(2,3,5)$ are not solutions and $(2,3,6)$ satisfies all the conditions. (We may observe $(2,4,5)$ is also not a solution.) Since $3+4+5=12>11=2+3+6$, it follows that $(2,3,6)$ has the required minimality.