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If $D$ is not too large, then it is surely the case that $\pi^D(N)$ goes to infinity when $N$ goes to infinity. So if you want to show that $\pi^2(N) \ge \pi^D(N)$, for all large $N$, then you also need to show that $\pi^2(N)$ goes to infinity with $N$. And this would resolve the twin prime conjecture. So no, I don't think that anyone can think of a way to show $\pi^2(N) \ge \pi^D(N)$ in that case (at this moment).

3 added 13 characters in body

If $D$ is not too large, then it is surely the case that $\pi^D(N)$ goes to infinity when $N$ goes to infinity. So if you want to show that $\pi^2(N) \ge \pi^D(N)$, for all large $N$, then you also show that $\pi^2(N)$ goes to infinity with $N$. And this would resolve the twin prime conjecture. So no, I don't think that anyone can think of a way to show $\pi^2(N) \ge \pi^D(N)$ in that case (at this moment).

2 deleted 13 characters in body

If $D$ is not too large, then it is surely the case that $\pi^D(N)$ goes to infinity when $N$ goes to infinity. So if you want to show that $\pi^2(N) \ge \pi^D(N)$, for all large $N$, then you also show that $\pi^2(N)$ goes to infinity with $N$. And this would resolve the twin prime conjecture. So no, I don't think that anyone can think of a way to show $\pi^2(N) \ge \pi^D(N)$ in that case (at this moment).

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