I am using Anthony Quas's reformulation to restate the problem. Letting the $n$th prime gap be given by $d_n= p_{n+1} - p_n$, and given $n$ we relabel $a=d_n$ and $b=d_{n+1}$, we look at $L$ as the set of $p_{n+1}$ in which $n$ satisfies $b-a\gt 0$ and we look at $Q$ as the set of $p_{n+1}$ in which $0 \lt (b-a)p_{n+1} -ab$. As noted in the post, $p_{n+1}$ not in $L$ readily implies $p_{n+1}$ not in $Q$; it is natural to ask if $p_{n+1} \in L$ implies $p_{n+1} \in Q$. The question further notes that when $p_{n+1}$ is observed to be in $L$ one also has $ ab - p_{n+1}(b-a) \gt (b-a)$, which I think should be reversed as $7=p_4 \in L$ but $49 - 55 \lt 11 - 14 + 5$.
If the last inequality is reversed, it says $ab \lt ( p_{n+1} + 1)(b-a)$. If $p_{n+1} \in Q$, then clearly this last inequality holds. Finally as Anthony Quas observes, if $p_{n+1} \in L\setminus Q$ then $ab \geq (b-a)p_{n+1}$ and $b \gt a$, so one would have $ab \geq 2p_{n+1}$ if $n \gt 1$.
The formulation shows that the basic question is about consecutive prime gaps, and that $L$ is different from $Q$ only when a large gap $d_n$ is greater than the square root of an adjacent prime $p_n$. Such large gaps have not been observed for $n \gt 30$ (so $p_n \gt 113$), and the stronger inequality $ab \geq 2p_{n+1}$ is also not observed for $1 \lt n \leq 30$. The case $n=1$ is left to the reader, as is the conclusion that $L$ properly containing $Q$ would violate expectations and many conjectures in prime number theory.
