Let $P(n)$ be the statement that $$n < \mathrm{rad}(n(n-1)(n-2)),$$ where $\mathrm{rad}$ is the radical of an integer, that is defined as $$\operatorname{rad}(m)=\prod_{\substack{p\mid m\\p\text{ prime}}}p$$ for integers $m>1$ with $\operatorname{rad}(1)=1$.
I checked that $P(n)$ holds for $3 \le n \le 3.10^7$.
My question: Is $P(n)$ true for any positive integer $n \geq 3$?
Also, is this a pre-existing conjecture?
Suppose we have such an $n$ where $P(n)$ is false. Now define $(a, b, c) := (1, n(n-2), (n-1)^2)$ and observe that these three numbers are pairwise coprime and satisfy $a + b = c$. Then we have:
$$ \textrm{rad}(abc) = \textrm{rad}(n(n-1)^2(n-2)) = \textrm{rad}(n(n-1)(n-2)) \leq n $$
Moreover, the radical of $n(n-1)(n-2)$ cannot be equal to any of $n$, $n-1$, and $n-2$ (because then that term would be squarefree, so equal to its own radical, and the other two terms in the product would make the radical strictly larger), so we actually have:
$$ \textrm{rad}(abc) \leq n - 3 < n - 1 = c^{\frac{1}{2}} $$
Such a counterexample $n$ would therefore result in an ABC triple with a quality $q > 2$, which is vastly larger than the current record-holder (which has a quality of 1.6299).
There are no such triples below $10^{20}$ (see here), and we can probably extend this much further by only looking for $q > 2$ instead of $q > 1.4$, but already this implies that your conjecture is true at least up to $10^{10}$.
