An inequality involving the radical of an integer and its greatest prime factor

Question

Let $n\geq 1$ an integer, in this post I denote the greatest prime dividing $n$ as $\operatorname{gpf}(n)$, and the product of the distinct prime numbers dividing $n$ as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p.$$

See it you want the Wikipedia Radical of an integer and the corresponding article from MathWorld Greatest Prime Factor.

In previous post of mine, see [1], I evoke an interesting inequality involving a previous arithmetic function.

Question (Updated, previous was too broad). I'm curious to know if it is possible to deduce an inequality of the form $$\operatorname{rad}(n)^{\alpha}<\text{something}\cdot\operatorname{gpf}(n)^{\beta},\tag{1}$$ that holds $\forall n>N$, for some integer $N$. Here $\text{something}$ means a function* of $n$; and $\alpha$ and $\beta$ is a suitable choice of positive real numbers. Many thanks.

*If you need it also depending on $\alpha$ and $\beta$, that is $\text{something}$ is a function of $n,\alpha$ and $\beta$, that is (see the feedback in comments) $$\text{something}=f(n,\alpha,\beta).$$

My example was added in the Appendix of the reference [1] this is my proposal $$\sqrt{\operatorname{rad}(n)}<\operatorname{gpf}(n)\cdot\log n$$ but seems that there are many counterexamples!

References:

[1] On an inequality involving the radical of an integer and its greatest prime factor, from this Mathematics Stack Exchange (2018).

Answer 1

It is clear that there will be expressions of that form that are satisfied for all $n$. In particular, consider $$ \text{rad}(n)<n\times\text{gpf}(n) $$ This is trivially true because $\text{rad}(n)<n$ and $\text{gpf}(n)>1$.

Note that, from here, I will consider $\beta=1$, as the choice is arbitrary (we can recover any other value by simply raising the inequality to the power of $\beta$).

We can place a bound on the "something" by looking at the case of primorials:

Consider $n=p\#$, the primorial for prime $p$. For this, we note that $\text{rad}(n)=n$ and $\text{gpf}(n)=p$. Now, we have $n^\alpha<f(n)\times p$.

However, it is known that $p\#\sim e^{(1+o(1))p}$ or $p\sim\frac{\ln p\#}{1+o(1)}$, and therefore we can observe that we require $$ n^\alpha<f(n)\times \frac{\ln n}{1+o(1)} $$ or, to put it another way, $$ f(n)>\frac{n^\alpha (1+o(1))}{\ln n} $$ Be careful with this, as what I have written isn't strictly correct, as the primorial only asymptotically approaches $e^{(1+o(1))p}$... but it gives a rough idea of the form for the best asymptotic "lower bound" for the "something".

Answer 2

I am skeptical that anything that is truly interesting can be deduced. Let $n=\prod_{i=1}^k p_i^{a_i}$ so that $p_i$ are primes and $a_i$ are positive integers. Then $\mathrm{gpf}(n)=p_k$. Your inequality then becomes $$ (p_1p_2\cdots p_k)^\alpha < s p_k^\beta $$ Where $s$ is a function of $\alpha,\beta,n$ and is your "something" in your question. Consider that from here, you construct a number $n'$ which has one more prime factor less than $p_k$, then if $\alpha$ is anything large (greater than $2$, really) then the LHS generally increases by a huge factor, while the RHS is constant. If in the worst case, we assume that the list $p_1,p_2,\ldots,p_k$ contains all the primes up to $p_k$, and we still can find an $s$ so that the inequality holds, then we can just add another prime, say, $p_{k+1}$ to the mix and create a new $n$ so that the LHS increases by a huge amount but the RHS only by a bit. So it seems as if the LHS is constantly trying to "break free" of the constraint of your inequality.

Of course you can easily counteract this by doing something like $\alpha=1$ and $s=n!$, but this result is hardly interesting at all.