With $N$ a constant $>0$, show $\prod_{p<x}\frac{1}{p^{N+1}-1}>\frac{0.2}{\log^2 x}$.

Related.

Show that if $x$ is large enough,$$\prod_{\substack{p<x \\ p \ \text{prime}}}\frac{1}{p^{N+1}-1}>\frac{0.2}{\log^2 x}.$$
Speaking of which, Theorem 6.12, and maybe others, of this paper might be useful.
If $N$ cannot be arbitrarily large for the inequality to hold, any conditions for truthfulness regarding its value are welcome.

Solutions Collecting From Web of "With $N$ a constant $>0$, show $\prod_{p<x}\frac{1}{p^{N+1}-1}>\frac{0.2}{\log^2 x}$."

Your inequality is equivalent to $$-\underset{p\leq x}{\sum}\log\left(p^{N+1}-1\right)>\log\left(0.2\right)-2\log\left(\log\left(x\right)\right).$$
Now we have, for partial summation and Prime Number Theorem, that exists $c_{1},c_{2}>0$
such that $$-\log\left(0.2\right)-\underset{p\leq x}{\sum}\log\left(p^{N+1}-1\right)<-\underset{p\leq x}{\sum}\log\left(p\right)=-\left(c_{1}x+c_{2}\frac{x}{\log\left(x\right)}+o\left(\frac{x}{\log\left(x\right)}\right)\right)<-cx<-2\log\left(\log\left(x\right)\right)$$
for some $c>0$
and $x$
large enough. So it is false.