= Solution
{c}
Write $D=u^a\nabla_a$, $\theta=\nabla_a u^a$ for the <expansion scalar>, $a_b=u^a\nabla_a u_b$ for the <four-acceleration>, and $P_b{}^c=\delta_b{}^c+u_bu^c$ for the <spatial projector of a timelike congruence>. The normalization of the <four-velocity> gives $u^ba_b=0$ and $u^b\nabla_a u_b=0$. The <product rule> and <metric compatibility> give the <covariant derivative> of the <perfect fluid in general relativity> <stress-energy tensor>:
$$
F_b:=\nabla^aT_{ab}
=u_b\bigl[D(\rho+p)+(\rho+p)\theta\bigr]+(\rho+p)a_b+\nabla_b p.
$$
Its <tensor contraction> with the <four-velocity> is
$$
u^bF_b=-D\rho-(\rho+p)\theta,
$$
and its projection with the <spatial projector of a timelike congruence> is
$$
P_b{}^cF_c=(\rho+p)a_b+P_b{}^c\nabla_c p.
$$
Thus <stress-energy conservation> implies the <relativistic perfect-fluid energy equation> and the <relativistic Euler equation>:
$$
\boxed{u^a\nabla_a\rho+(\rho+p)\nabla_a u^a=0,\qquad
(\rho+p)u^a\nabla_a u_b+(g_{ab}+u_au_b)\nabla^a p=0.}
$$
Conversely, the two displayed equations set both projections of $F_b$ to zero. Since $F_b=P_b{}^cF_c-u_bu^cF_c$, they imply $F_b=0$. This establishes equivalence, including points where $\rho+p=0$, without dividing by that quantity.
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