The title says it all:
Is it true or can it be proven that $\frac{n}{\varphi(n)} \geq \operatorname{rad}(n)$?
Here $n$ is a natural number, $\varphi(n)$ is the Euler-totient function of $n$, and $\operatorname{rad}(n)$ is the radical of $n$.
Here is my:
ATTEMPT AT A PROOF
We know that $$\varphi(n) = n\prod_{p \mid n}\bigg(1 - \frac{1}{p}\bigg),$$ so that we obtain $$\frac{n}{\varphi(n)} = \prod_{p \mid n}\bigg(\frac{p}{p-1}\bigg) = \frac{\bigg(\prod_{p \mid n}{p}\bigg)}{\bigg(\prod_{p \mid n}(p-1)\bigg)} = \frac{\operatorname{rad}(n)}{\bigg(\prod_{p \mid n}(p-1)\bigg)}.$$
Since the $p$'s are primes, then $p \geq 2$. Thus equality holds only when $n$ is a power of two.
Otherwise, $$\frac{n}{\varphi(n)} < \operatorname{rad}(n),$$ since then the $$\bigg(\prod_{p \mid n}(p-1)\bigg)$$ term in the denominator would be greater than one.
We therefore conclude that, in general, $$\frac{n}{\varphi(n)} \leq \operatorname{rad}(n)$$ holds, with equality occurring only if $n$ is a power of two. (Note that $n=2^0=1$ is included.)
QUESTION
Does this (dis)proof suffice?