Question

Let $a$ and $b$ be positive integers satisfying $\frac{ab + 1}{a + b} \lt \frac{3}{2}.$ The maximum possible value of $\frac{a^3 b^3 + 1}{a^3 + b^3}$ is $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

Answer

Solution:

If $a = 1$ or $b = 1,$ then $\frac{a^3b^3 + 1}{a^3 + b^3} = 1.$ So assume $a, b \ge 2.$ Clearing denominators, the hypothesis says $2ab + 2 \lt 3a + 3b,$ and multiplying by $2$ and rearranging gives $(2a - 3)(2b - 3) = 4ab - 6a - 6b + 9 \lt 5.$

For $a, b \ge 2$ both factors are positive odd integers, so up to symmetry the only options are $(a, b) = (2, 2)$ and $(2, 3)$ (both of which do satisfy the original inequality, while $(3, 3)$ gives the product $9$ ).

The values are $\frac{65}{16}$ for $(2, 2)$ and $\frac{8 \cdot 27 + 1}{8 + 27} = \frac{217}{35} = \frac{31}{5}$ for $(2, 3).$ The larger is $\frac{31}{5},$ so $p + q = 31 + 5 = 36.$

2015 AIME II Problem 8

Solution:

Problem 8 in Other Years