Following from the asymptotic relations described here,
I am seeking a proof or minimal counter example for the inequality below, the right hand side of which occurs for $n=9$ and $k=2$.
$$n-\Biggl(n^k -{\lfloor n^{\frac{1}{k}} \rfloor}^{k-1}\gcd\Bigl({\lfloor n^{\frac{1}{k}} \rfloor},\Bigl\lfloor \frac{p_n^{k}}{n^{k}} \Bigr\rfloor\Bigr)\Biggr)^{\frac{1}{k}} \leq 9 - \sqrt{72} \quad \forall {\{n,k}\} \subset \mathbb N\backslash {\{1}\}\tag{0}$$
I have tested the conjecture up to $\max({\{n,k}\}) \leq 100$ but the computations require a lot of processing power for large $k$.
Further context asks as to whether or not the following inequality holds, the right hand side for this case being the value of the left hand side when $n=2$ and $k=2$: $$\frac{\Biggl(n^k -{\lfloor n^{\frac{1}{k}} \rfloor}^{k-1}\gcd\Bigl({\lfloor n^{\frac{1}{k}} \rfloor},\Bigl\lfloor \frac{p_n^{k}}{n^{k}} \Bigr\rfloor\Bigr)\Biggr)^{\frac{1}{k}}}{n} \geq \frac{\sqrt{3}}{2} \quad \forall {\{n,k}\} \subset \mathbb N\backslash {\{1}\}\tag{1}$$ Again for further context, particular for first relation stated in the linked content, as with $(0)$ the boundary value occurs for $n=9$ and $k=2$:
$$\Biggl|1-\Biggl(\frac{n^k -{\lfloor n^{\frac{1}{k}} \rfloor}^{k-1}\gcd\Bigl({\lfloor n^{\frac{1}{k}} \rfloor}^{k-1},\Bigl\lfloor \frac{p_n^{k-1}}{n^{k-1}} \Bigr\rfloor\Bigr)}{n^k -{\lfloor n^{\frac{1}{k}} \rfloor}\gcd\Bigl({\lfloor n^{\frac{1}{k}} \rfloor},\Bigl\lfloor \frac{p_n^{k}}{n^{k}} \Bigr\rfloor\Bigr)}\Biggr)^{\frac{1}{k}}\Biggr| \leq \frac{\sqrt{3}\sqrt{13}}{6}-1\tag{2}$$
Thankyou for your time.
