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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).

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  • $\begingroup$ Honestly I think the whole idea is a bit hard to understand, and the question might seem specific enough, but in some sense it is quite broad: the ‘something’ is too general. $\endgroup$ – Szeto Jun 27 '18 at 12:24
  • $\begingroup$ Also, why would you want to ‘create’ such inequality? $\endgroup$ – Szeto Jun 27 '18 at 12:25
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    $\begingroup$ Yes, I understand you want something valuable, which my example isn't, even though it matches your requirements. This is a very common situation in math: you have a vague perception of some beautiful idea, but you can't quite pinpoint it, you try this way and that, and when you succeed, it vanishes. Maybe it never was there in the first place. Who knows? $\endgroup$ – Ivan Neretin Jun 27 '18 at 16:54
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    $\begingroup$ In the worst case, $n=p_k\#$ for some prime $p$, $\operatorname{rad}(n)=n$ and $\operatorname{gpf}(n)=p_k$. Since $n \approx e^k$, your "something" should be at least exponential. $\endgroup$ – didgogns Jun 28 '18 at 12:40
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    $\begingroup$ If alpha,beta and something are constants, it is trivially not possible just by considering the product of first $n$ primes. $\endgroup$ – Haran Jul 14 '18 at 7:05
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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".

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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.

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