Assume p £ q £ r. In order to maximise the reciprocals we want to make p, q and r as small as possible. If p ³ 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 £ 1/2.

If q = 4 then r ³ 5 (because 1/4 + 1/4 = 1/2) so 1/p + 1/q + 1/r £ 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.