Anemone and kaliprasad have noticed that nobody ever replied to this challenge problem. Here is my attempt, using MarkFL's favourite method of Lagrange multipliers.
To find the extreme points of $abc$ subject to the restraint $bc+ca+ab = k$ (where $\frac13\leqslant k\leqslant 3$), put the partial derivatives of $abc - \lambda(bc+ca+ab - k)$ (with respect to $a$, $b$ and $c$) equal to $0$: $$bc - \lambda (b+c) = 0,\qquad ca - \lambda (c+a) = 0,\qquad ab - \lambda (a+b) = 0.$$ Write those equations as $$\frac1\lambda = \frac1b + \frac1c = \frac1c + \frac1a = \frac1a + \frac1b$$ to see that $a=b=c$. That is the unique extremal point of $abc$. It must be a maximum because if we take $b=c=\varepsilon$ and $a = \dfrac{k-\varepsilon^2}{2\varepsilon}$ then $bc+ca+ab = k$ but $abc = \frac12\varepsilon(k-\varepsilon^2) \to0$ as $\varepsilon\to0$. So $abc\to0$ towards the boundary of the set $\{(a,b,c)\in \mathbb{R}^3:a>0,\,b>0,\,c>0\}.$ Thus the maximum possible value of $abc$ occurs when $k=3$ and $a=b=c= abc =1$. The range of values of $abc$ is therefore the half-open interval $(0,1]$.
An exactly similar calculation for the sum $a+b+c$ shows that it can take arbitrarily large values (when $b=c= \varepsilon$, $a = \dfrac{k-\varepsilon^2}{2\varepsilon}$ and $\varepsilon\to0$). There is again a unique extremal point when $a=b=c$, but this time it is a minimum, occurring when $a=b=c=\frac13$ and $a+b+c=1$. So the range of values of $a+b+c$ is the interval $[1,\infty).$
