Seems to me that for 3 consecutive primes $p_1, p_2, p_3, p_1>2$, it is always the case that $p_1+p_2 > p_3$. Do you know a proof?
(Initially I thought there must be large primes for which this is the case, but the limit as $n \to \infty$ of the ratio of $n^{th}$ prime gap to the nth prime is 0...)
If not, then there exist consecutive primes $p$ and $q$ for which there is no prime in the set
$$\{q+1,q+2,\cdots,q+p\}.$$
This is almost Bertrand's postulate, but not quite. Per this page, it has been proven that, for all $n\geq 25$, there is a prime in the range
$$(n,6n/5),$$
which means there is one in the range
$$(q,6q/5).$$
So, if $6q/5\leq q+p$, we're done for all $q\geq 25$ (the others can be checked manually) Indeed, this is equivalent to $q/5\leq p$ or $q\leq 5p$. However, there's a prime in the range $(p,2p)$, which means $q\leq 2p < 5p$, so we're done.
