Sum of two reciprocals

using $Mathematica$

{m, n}/.Solve[1/n+1/m>=.6&&1<n<=m,{n,m},Integers]

returns {{2, 2}, {3, 2}, {4, 2}, {5, 2}, {6, 2}, {7, 2}, {8, 2}, {9, 2}, {10, 2}, {3, 3}}

Given that n is smaller or equal to m but bigger than 1, 1/n must be at least 0.3 and therefore n must be either 2 or 3 as 1/4 is smaller than 0.3 (it is 0.25) but 1/3 is bigger (it is 0.33333...). When n=2, the new equation is 1/2 + 1/m ≥ 0.6 and therefore 1/m ≥ 0.1=0.6-0.5. 1/m = 0.1 when m=10 so m can be any value smaller than 10 up to 2 when n is 2. We now have 9 pairs: (n,m)=(2,2) or (2,3) or (2,4) or (2,5) or (2,6) or (2,7) or (2,8) or (2,9) or (2,10).

However, when n=3, m must be 3 as 1/3 +1/4 is less than 0.6 (it is 0.58333...) but 1/3 +1/3 is bigger than 0.6 (it is 0.666666...). We now have one more pair: (3,3).

Therefore overall there are 9+1= 10 pairs . These being (2,2), (2,3), (2,4), (2,5), (2,6), (2,7), (2,8), (2,9), (2,10) and (3,3).