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.

*I'm interested in create interesting inequalities involving previous arithmetic functions, with a good mathematical content. After some experiments with these arithmetic functions I wondered next question.

Question. Are there infinitely many integers $n\geq 1$ such that $$\sqrt{\frac{\operatorname{rad}(n)}{2}}>\operatorname{gpf}(n)\tag{1}$$ holds? Many thanks.

The sequence of the first few integers satisfying $(1)$ starts as


and I suspect that should be infinitely many of them. I don't know how to solve it, maybe using inequalities about different means, and invoking some theorem about prime numbers.


*I did more experiments combining these arithmetic functions, similar than previous or for example this different inequality for integers $n>1$

$$\frac{\sqrt{\operatorname{rad}(n)}}{\log n}>\operatorname{gpf}(n) ,$$

the first few solutions of this last inequality are $2,30030,34034,36465,39270,40755\ldots$

Thus the corresponding inequalities $<$ to $(1)$ and this last inequality seem that are satisfied for many much positive integers.


For the first part, take any primes $p, q$ with $4 < p < q$. By Bertrand's postulate (proved by Chebyshev) there exists a prime $r$ such that $2q > r > q$. Let $n = pqr$, then ${\rm rad}(n) = pqr$ and ${\rm gpf}(n) = r$. The desired inequality $\sqrt(pqr/2) > r$ is equivalent to $2pq > 4r$, which follows because $p > 4$ and $2q > r$.

  • $\begingroup$ First a clarification: there is only a Question in my post. And I believe that you prove it, thus as soon as I understand your proof I am going to accept your answer. Many thanks. $\endgroup$ – user243301 Jun 27 '18 at 11:25

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