If $$p^{r + 1} - 1 > 5(p^r - 1),$$ then is there a lower bound for $p$ (e.g., in terms of $r$) and $r$ (e.g., in terms of $p$)? Here, $p>1$ and $r \geq 1$.
I tried using WolframAlpha, but I do not understand the resulting inequality plots.
I also tried the Reduce built-in function in Mathematica (for which InequalitySolve is a precursor), but it returns the error message Reduce::nsmet: This system cannot be solved with the methods available to Reduce.
Note this does not actually answer the question on WolframAlpha but attempts to algebraically solve the inequality. My only thought on that is that the shaded region is the solution set. For example, the coordinate $(r,p)=(0,2)$ is in the shaded region, thus $2^1-1>5(2^0-1)$.
When $p^r>1,p>1$, we have
$$\frac{p^{r+1}-1}{p^r-1}=\frac{p^r+p^{r-1}+\dots+1}{p^{r-1}+\dots+1}=1+\frac{p^r}{p^{r-1}+\dots+1}>5$$
Thus,
$$p^r>4(p^{r-1}+\dots+1)>4p^{r-1}+4\int_0^{r-1}p^x\ dx=4p^{r-1}(1+\ln(p))-4>4p^{r-1}\ln(p)$$
Thus, we find that
$$p>4\ln(p)$$
$$p>8.613$$
These are some solutions that hold for any $r$ such that $p^r>1,p>1$ holds.
Without so much technical stuff, one could just note that when $p\ge5$, we have
$$p^{r+1}-1>p^{r+1}-5\ge5\times p^r-5=5(p^r-1)$$
Thus it holds for all $p\ge5$.