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!
 A: Try manipulating the first inequality to define $a$ in terms of $b$ (or vice versa)
$$ \frac{ab+1}{a+b} < \frac{3}{2} \Rightarrow 2ab+2 < 3a+3b$$ 
$$ \Rightarrow 2ab - 3a < 3b-2 $$ 
$$ \Rightarrow a < \frac{3b-2}{2b-3}$$
Notice that if $a=1$, then the second fraction involving $a$ and $b$ would become
$$ \frac{(1)^3b^3+1}{(1)^3+b^3} = \frac{b^3+1}{b^3+1} = 1$$
Similarly, if $b=1$, this second fraction would become
$$ \frac{a^3(1)^3+1}{a^3+(1)^3} = \frac{a^3+1}{a^3+1} = 1$$
Let's exclude these possibilities for $a$ and $b$ because they don't offer much useful information for finding a maximum bound. Then since $a$ and $b$ are both positive integers, we can setup a new inequailty for $a > 2$ using the value of $a$ we defined in terms of $b$
$$ \frac{3b-2}{2b-3} > 2$$
Since we want $b > 1$, we can multiply both sides by $2b-3$ to simplify
\begin{align*}
\frac{3b-2}{2b-3}\cdot (2b-3) &> 2\cdot(2b-3)\\
3b-2 &> 4b-6\\
\Rightarrow b < 4
\end{align*}
Since you said "Hints only" I'll leave the rest up to you. Let me know if you'd like further help!
A: 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 $(a,b)\in\big\{(2,2),(2,3)\big\}$.
