[Max of $abc$ with fixed $e_1, e_2$]

The numbers $a, b, c$ are the roots of the cubic

$

t^{3} - 6t^{2} + 9t - p = 0, \quad \text{where } p = abc.

$

Define $g(t) = t(t-3)^{2} = t^{3} - 6t^{2} + 9t$. The cubic becomes $g(t) = p$. We need three real non-negative roots.

$g'(t) = 3(t-1)(t-3)$: local max $g(1) = 4$, local min $g(3) = 0$. For $0 \le p \le 4$, $g(t) = p$ has three real roots. For $0 < p < 4$, all three roots lie in $(0, \infty)$ (one in $(0,1)$, one in $(1,3)$, one in $(3, \infty)$). At the boundary $p = 4$, the roots are $1, 1, 4$, which satisfy $1+1+4 = 6$ and $1\!\cdot\!1 + 1\!\cdot\!4 + 1\!\cdot\!4 = 9$. For $p > 4$, only one root is real.

Hence the maximum of $abc$ is $4$, attained at $(a, b, c) = (1, 1, 4)$ (and permutations). Answer: (C) 4.

Why CAT-level: The Newton-Vieta linkage between $abc$ and a cubic in one variable is non-obvious; AM-GM gives only weak bounds here.