inequality

Obviously, p,q,r > 1
(1/p)+(1/q)+(1/r) obtains maximum when p,q and r are of the smallest possible values.
However, they cannot be all 2 at the same time.
So, when one of p,q and r is 2, the smallese possible value of the other two is 3.
When one of them is 2 and one of them is 3:
if the last one is 4, (1/2)+(1/3)+(1/4) = 13/12 ＞ 1
5, (1/2)+(1/3)+(1/5) = 31/30 ＞ 1
6, (1/2)+(1/3)+(1/4) = 1
7, (1/2)+(1/3)+(1/7) = 41/42＜ 1
For the last one ＞ 7, the value of (1/p)+(1/q)+(1/r) ＜ 41/42
Hence, 41/42 is the maximum possible value foe (1/p)+(1/q)+(1/r) if (1/p)+(1/q)+(1/r) ＜ 1.
i.e. (1/p)+(1/q)+(1/r) ＜= 41/42

(I'm an S.4 student only. I don't know the formal proof)