Triples of Numbers

Well, $\frac{1}{3} + \frac{1}{4} + \frac{1}{5} < 1$, so we must have $a=2$. So we really just need $\frac{1}{b} + \frac{1}{c} = \frac{1}{2}$.

Since $\frac{1}{4} + \frac{1}{5} < \frac{1}{2}$, $b = 3$. That leaves $c = 6$.

I think a nice way to think about it is to view the number $1$ as $\frac{1}{1}$. We can decompose $\frac{1}{n}$ into $\frac{1}{n+1} + \frac{1}{n^2+n}$, then decompose one of those to get an expression for $\frac{1}{n}$ as the sum of three harmonic numbers. In this case, we see that $\frac{1}{1} = \frac{1}{2} + \frac{1}{2} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$.

See the Leibniz Harmonic Triangle.

HINT $$ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \lt \frac{1}{a} + \frac{1}{a} + \frac{1}{a}, $$ which shows that $a \leq \ldots$.¹

(After fixing $a$, you can use the same idea again to complete the proof.)

¹EDIT: Corrected the first inequality sign from $\gt$ to $\lt$.