You've shown that one set of $14$ relatively prime and non-prime values cannot be extended to $15.$ But that doesn't show it for all sets of $14$ non-prime relatively prime values. Another $n=14$ example set is $\{2p_{27},3p_{26},5p_{25},\cdots,41p_{15},43^2\},$ where $p_{15}=47,p_{16}=53,\dots,p_{27}=103.$
Your instinct is correct, but it is just not a complete proof.
This is a neat problem, because the intuition feels instinctive, but proving it correctly requires some technique.
Given any $m>1$, define $d(m)$ to be the smallest prime divisor of $m.$
Claim 1: If $2\leq m\leq 2004$ is not prime, $d(m)\leq 43.$
Proof: Otherwise, $m\geq p_1p_2$ for some pair of primes $p_1,p_2$ and $p_i\geq 47.$ But then $m\geq 47^2=2209.$
Claim 2: If $m_1,m_2>1$ are relatively prime, then $d(m_1)\neq d(m_2).$
Proof: If $p=d(m_1)=d(m_2)$ then $p$ us a common factor of $m_1$ and $m_2.$
Claim 3: Given any set $S=\{m_1,\cdots,m_{15}\}$ of non-prime values with $2\leq m_i\leq 2004.$ Then the set is not pairwise relatively prime
Proof: There are at most $14$ distinct possible values of $d(m_i)$ by claim $1,$ since there are $14$ distinct primes $\leq 43$. Thus $d(m_i)=d(m_j)$ for some $i\neq j.$ Then by claim $2,$ $m_i$ and $m_j$ are not relatively prime.
Your example of the squares $S=\{2^2,\cdots,43^2\}$ shows that $n=15$ is the smallest such $n$ for which it is true.
Applying the AM-GM inequality for 3 times $\frac{x}{3}$, 4 times $\frac{y}{4}$ and 5 times $\frac{z}{5}$, we have
\begin{align}
12 &= x+y+z \\ & = 3\times \frac{x}{3} + 4 \times \frac{y}{4} +5 \times \frac{z}{5} \ge (3+4+5)\sqrt[12]{\left(\frac{x}{3}\right)^3\left(\frac{y}{4}\right)^4\left(\frac{z}{5}\right)^5} = 12
\end{align}
The equality occurs if and only if $\frac{x}{3} = \frac{y}{4}= \frac{z}{5}$ and $x+y+z= 12$ or
$$(x,y,z) = (3,4,5)$$
Q.E.D
Best Answer
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 $(a,b)\in\big\{(2,2),(2,3)\big\}$.