= Solution
For $\mu>0$, the <logarithmic barrier> objective is
$$
\boxed{f_\mu(x)=f(x)-\mu\sum_{i=1}^p\log c_i(x)},\qquad D=\{x:c_i(x)>0\text{ for all }i\}.
$$
The <domain of a function> $D$ is a nonempty <convex set> by the same <concavity> argument as in part (a), with strict inequalities. Differentiating twice on $D$ gives
$$
\nabla^2f_\mu(x)=\nabla^2f(x)-\mu\sum_i\frac{\nabla^2c_i(x)}{c_i(x)}+\mu\sum_i\frac{\nabla c_i(x)\nabla c_i(x)^T}{c_i(x)^2}.
$$
Here $\nabla^2f\succeq0$ by <convexity>, $-\nabla^2c_i\succeq0$ by <concavity>, and each <outer product> is a <positive semidefinite matrix>. Hence $\nabla^2f_\mu\succeq0$, proving <convexity> on $D$.
The domain qualification matters: the displayed real-valued <logarithm> expression is not defined on all of $\mathbb R^n$. The precise whole-space version is the extended-real <convex function> equal to $f_\mu$ on $D$ and $+\infty$ outside $D$.
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