Solution:

Let $g$ and $l$ denote the greatest common divisor and the least common multiple, respectively, of $a$ and $b$. Then $g l = a b$. Therefore $g + l \leq a b + 1$.

Suppose that $(g + l) / (a + b) > (a + b) / 4$. Then we have $a b + 1 > (a + b)^2 / 4$, so we get $(a - b)^2 < 4$.

Assuming $a \geq b$ we either have $a = b$ or $a = b + 1$.

In the former case, $g = l = a$ and the quotient is $(g + l) / (a + b) = 1 \leq (a + b) / 4$.

In the latter case, $g = 1$ and $l = b(b + 1)$ so we get that $2b + 1$ divides $b^2 + b + 1$. Therefore $2b + 1$ divides $4(b^2 + b + 1) - (2b + 1)^2 = 3$ which implies that $b = 1$ and $a = 2$, a contradiction to the given assumption that $a b > 2$.

This shows that $(g + l) / (a + b) \leq (a + b) / 4$.

Note that for the equality to hold, we need that either $a = b = 2$ or, $(a - b)^2 = 4$ and $g = 1$, $l = a b$. The latter case happens if and only if $a$ and $b$ are two consecutive odd numbers. (If $a = 2k + 1$ and $b = 2k - 1$ then $a + b = 4k$ divides $a b + 1 = 4k^2$ and the quotient is precisely $(a + b) / 4$.)