Note that

1 / 3 + 1 / 3 + 1 / 4 = 11 / 12

1 / 2 + 1 / 3 + 1 / 6 = 1

and

1 / 2 + 1 / 3 + 1 / 7 = 41 / 42

Assume $p \leq q \leq r$ . In order to maximise the reciprocals we want to make $p$ , $q$ and $r$ as small as possible. If $p \geq 3$ then we can't have $p = q = r = 3$ as that would give a total of 1, so the maximum with $p = 3$ is $p = q = 3$ and $r = 4$ which is $11 / 12$ and is not as big as $41 / 42$ .

So for the maximum $p = 2$ and we are looking for $q$ and $r$ such that $1 / q + 1 / r \leq 1 / 2$ .

If $q = 4$ then $r \geq 5$ (because $1 / 4 + 1 / 4 = 1 / 2$ ) so $1 / p + 1 / q + 1 / r \leq 1 / 2 + 1 / 4 + 1 / 5 = 19 / 20 < 41 / 42$ .

So $q = 3$ . Hence $1 / p + 1 / q + 1 / r = 1 / 2 + 1 / 3 + 1 / r < 1$ and this equals 1 if $r = 6$ so the maximum value must occur when $r = 7$ and it must be $41 / 42$ .