, 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 $a \neq b$. 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.