Is there a lower bound for $\operatorname{rad}(n)$ in terms of $n \in \mathbb{N}$?

Question

For integers $n\geq 1$ we denote the square-free kernel as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ that is the product of distinct primes dividing an integer $n>1$ with the definition $\operatorname{rad}(1)=1$ (the Wikipedia's article dedicated to this multiplicative function is Radical of an integer).

Inspired by this earlier question, here is my own question:

Is there a lower bound for $\operatorname{rad}(n)$ in terms of $n \in \mathbb{N}$?

Sure, I know that $\operatorname{rad}(n)$ is (weakly) multiplicative, and that $\operatorname{rad}(1)=1$, but apart from these two pieces of information, I cannot seem to figure out how to obtain a lower bound, however trivial.

In particular, might it be possible to prove that $$\operatorname{rad}(n) \geq \omega(n)$$ (where $\omega(n)$ is the number of distinct prime factors of $n$), or even $$\operatorname{rad}(n) \geq \Omega(n)$$ (where $\Omega(n)$ is the number of prime factors of $n$ counting multiplicity)?

Updated after the answers The current known (somewhat trivial) lower bound is $$\operatorname{rad}(n) > 2^{\omega(n)}.$$

Answer 1

We have

$$\operatorname{rad} n = \prod_{p \mid n} p \geqslant \prod_{k = 1}^{\omega(n)} p_k\,.$$

Using $p_k \geqslant k+1$ we obtain the bound $\operatorname{rad} n \geqslant (\omega(n)+1)!$. Using that all primes larger than $2$ are odd, we obtain

$$\operatorname{rad} n \geqslant 2\cdot \prod_{k = 2}^{\omega(n)} (2k-1) = 2\cdot (2\omega(n)-1)!! = \frac{(2\omega(n))!}{2^{\omega(n)-1}\omega(n)!}$$

for $n > 1$, and by Stirling's formula that lower bound is asymptotically equal (for $\omega(n) \to \infty$) to

$$\frac{1}{2^{\omega(n)-1}}\sqrt{\frac{2\pi 2\omega(n)}{2\pi\omega(n)}}\biggl(\frac{2\omega(n)}{e}\biggr)^{2\omega(n)}\biggl(\frac{e}{\omega(n)}\biggr)^{\omega(n)} = 2^{\omega(n) + \frac{3}{2}}\biggl(\frac{\omega(n)}{e}\biggr)^{\omega(n)}.$$

And using the prime number theorem we get an asymptotic lower bound via

$$\log \operatorname{rad} n \geqslant \sum_{k = 1}^{\omega(n)} \log p_k = \vartheta(p_{\omega(n)}) \sim p_{\omega(n)} \sim \omega(n)\log \omega(n)\,,$$

so anything growing sufficiently slower than $\omega(n)^{\omega(n)}$ can serve as a lower bound with a suitable factor to capture the cases where $\omega(n)$ is small. And anything growing sufficiently faster can't be a lower bound. [The weaselly "sufficiently" can be made more precise with more precise estimates for $\vartheta(x)$ and $p_n$.]

However, I doubt the practical usefulness of these bounds. If you know $\omega(n)$, then you typically know enough about $n$ to have much sharper estimates for $\operatorname{rad} n$.

Answer 2

Blimey, how could I have missed it!

Since each prime $p$ in the factorization of $n$ is at least $2$, and since $n$ has $\omega(n)$ distinct prime factors, then a lower bound for $\operatorname{rad}(n)$ is $$2^{\omega(n)}.$$

Perhaps this (very) trivial lower bound could be improved?

ADDED AFTER INITIAL ATTEMPT

I am thinking this lower bound could be improved to $$\bigg(n!\bigg)^{\omega(n)},$$ but I currently do not have a complete proof. (Edit June 9 2018: As pointed out by B. Goddard in the comments, this second inequality fails for $n=5$.)

Answer 3

Well, you have $\omega(n)$ prime factors in rad$(n),$ each of which is at least $2$, so you have

$$\mbox{rad}(n) >2^{\omega(n)}. $$