Solution:

For fixed denominators $a < b$ (with sum $10$), we maximize the sum of the fractions by giving the smaller denominator as large a numerator as possible: $8/a + 1/b$. Then, if $a \geq 2$, this quantity is at most $8/2 + 1/1 = 5$, which is clearly smaller than the sum we get by setting $a = 1$, namely $8/1 + 1/9 = 73/9$. So this is the answer.