Question in the title.
It intuitively seems absurd that $p_N - p_{N-1} \gt p_{N-1} - 3 = $ the largest gap formable from all $p_i = $ odd primes $3, \dots, p_{N-1}$.
Was wondering how difficult the proof is.
$2 p_i$ is the smallest composite divisible by $p_i$. And $p_N + 3$ is certainly a composite. Not sure if that helps : >
Thanks @PaoloLeonetti in the comments. According to the article:
In 1998, Pierre Dusart improved the result in his doctoral thesis, showing that for $k \geq 463, p_{k+1} \leq (1 + 1/(\ln^2 p_k))p_k$, ...
So we want to show that $p_{k+1} \leq 2 p_{k} - 3$ for sufficiently large $k$.
$2 p_k - 3 = (2 - \dfrac{3}{p_k})p_k$ and
$$ 2 - \dfrac{3}{p_k} \geq 1 + 1 / (\ln^2 p_k) \iff (1 - \dfrac{3}{p_k})\ln^2 p_k \geq 1 \iff \\ \ln^2 p_k \geq \dfrac{p_k}{p_k - 3} $$
the last operation being valid since $k \geq 463$ and so $3$ is much less than $p_k$ and so $1 - \dfrac{3}{p_k} \gt 0$.
Now take the exponential:
$$ \iff 8 \approx \ln(p_k) \geq e^{\frac{p_k}{p_k - 3}} \approx 2 $$ Where the approximation is at least valid enough for $k \geq 463$.
It is known that if $n\ge 25$, there is always a prime $p$ such that $n\le p\le 6n/5$ (Jitsuro Nagura, 1952). On the other hand, your claim is $$ p_{k+1} \le 2p_k-3 $$ for all $k$. Thanks to the above result, if $p_k \ge 29$, there exists a prime $p^\star$ in the interval $[p_k+1, \frac{6}{5}(p_k+1)]$. Moreover, $p^\star$ will be at least $p_{k+1}$, so $$ p_k < p_{k+1} \le p^\star \le \frac{6}{5}(p_k+1) $$ and it will be sufficient to prove that $\frac{6}{5}(p_k+1) \le 2p_k-3$, i.e., $p_k \ge \frac{21}{4}$. And this is verified if $p_k \ge 29$. For the remaining cases, check it manually.