<s>[INST] Find the maximum value of the fraction 
Let $a$ and $b$ be positive integers satisfying $\frac{ab+1}{a+b}<\frac{3}{2}$. The maximum possible value of $\frac{a^3b^3+1}{a^3+b^3}$ is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Trial and error makes the job very easy, but it isn't rigorous. 
I used factoring:
$$= \frac{(ab + 1)(a^2b^2 - ab + 1)}{(a+b)(a^2 - ab + b^2)} < \frac{3}{2} \cdot \bigg( \frac{a^2b^2 -ab + 1}{a^2 -ab + b^2} \bigg)$$
But that doesnt get you anywhere either.
Hints only please!
 [/INST] Without loss of generality, we assume that $a\leq b$.  Since $a=2\left(\frac{ab}{2b}\right)\leq2\left(\frac{ab}{a+b}\right)< 2\left(\frac{ab+1}{a+b}\right)<3$, we have $a=1$ or $a=2$.  If $a=1$, then $b$ can be any natural number and $\frac{a^3b^3+1}{a^3+b^3}=1$.  If $a=2$, then $\frac{2b+1}{b+2}=\frac{ab+1}{a+b}<\frac{3}{2}$ gives $4b+2<3b+6$, or $b<4$.  The rest is only checking with $(a,b)\in\big\{(2,2),(2,3)\big\}$.
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