Is this set dense in [0,+∞)? We define $A=\{ \frac{c}{rad(abc)}: a, b > 0, c=a+b, gcd(a, b)=1 \}$.
Is the set $A$ dense in $[0, +\infty)$?
Does $\overline{A}$ have interior? Here $\overline{A}$ is the closure of $A$.
A well-known fact is that $\inf A=0$ and $\sup A=+\infty$.
 A: This is not a full answer, but a pair of soft arguments suggesting that $A$ is dense in $[0, +\infty)$.

First Argument
Given any triple $(a,b,c)$, let $\displaystyle r(a,b,c)=\frac{c}{\text{rad}(abc)}$.
One can generate two new triples
$$t_1=(a(c+b),b^2,c^2)$$
$$t_2=(a^2,b(c+a),c^2)$$
with ratios
$$r(t_1)=r(a,b,c)\cdot\frac{c}{c+b}\cdot\frac{c+b}{\text{rad}(c+b)}$$
$$r(t_2)=r(a,b,c)\cdot\frac{c}{c+a}\cdot\frac{c+a}{\text{rad}(c+a)}$$
Clearly $\displaystyle\frac{1}{2}<\frac{c}{c+a}, \frac{c}{c+b}<1$.
On the other hand $\displaystyle \frac{n}{\text{rad}(n)}$ is on multiplicative average equal to $\prod_{p} p^{1/(p^2-p)}\approx2.128$. So the tree of triples generated by repeated application of the transformations above,  will form paths of ratios that drift (mostly slowly) to increasingly higher values.
$0$ being an obvious accumulation point of $A$, one has infinitely many starting points for the process above, making it plausible that any presumed gap in $A$ will be cut by infinitely many paths.

Second Argument
This is via an example specifically targeting the gap around $0.87$ that was mentioned in a comment to the question. Start by approximating $0.87$ with a rational with squarefree denominator (and preferably multiple small prime factors). $61/70$ is a good candidate. Take the triple $(1, 2^{100}\cdot 5^7\cdot 7^4, 2^{100}\cdot 5^7\cdot 7^4+1)$ where the exponents $100,7,4$ are picked so that $61^2$ divides $c$. There is then a good chance that $c/\text{rad}(c)=61$, which is indeed the case here, and the ratio of the triple is therefore $61/70$.
