a b c, it's as easy as 1 2 3

First note that we must have $\frac 1a < 1$ , so $a > 1$ .

Since $\frac 1a > \frac 1b > \frac 1c$ , we must also have $\frac 1a > \frac 13$ ; so $a < 3.$ Thus, $a = 2$ .

Now $\frac 1b + \frac 1c = \frac 12$ where $2 < b < c.$

Similar to before, $\frac 1b > \frac 14$ , so $b < 4.$ Thus, $b = 3.$

With $a = 2$ and $b = 3$ , we have $\frac 12 + \frac 13 + \frac 1c = 1,$ which is satisfied when $c = 6.$

To conclude, $a + b + c = 2 + 3 + 6 = 11.$