How to solve this ???

@above: The key restriction to this question is that $a,b,c$ are positive integers. Apply some upper bounds and you'll eventually find that "equality"; and from there, the rest will basically just be fill in.

Notice that $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}=\frac{41}{42}$. Without loss of generality, suppose $a\leq b\leq c$. If $a=2$ and $b=3$, then keeping in mind the first condition, $c=7$ is optimal, befitting what we want to prove. If $a=2$, $b\geq 4$ and $c \geq 5$, we get a bound weaker than the desired bound. If $a \geq 3$, $\, b \geq 3$ and $c\geq 4$, we again get a weaker bound. Hence, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq \frac{41}{42}$ and equality is achieved at $\left(2, 3, 7 \right)$ and its permutations.