= Solution
Write $A=\sigma\sigma^T$ and let $V(x,t)$ denote the optimal remaining expected reward from positive <portfolio wealth> $x$ at time $t$. The <Hamilton-Jacobi-Bellman equation> and terminal condition are
$$
\boxed{V_t+\sup_{\theta\in\mathbb R^d,\ c\geq0}\left\{(\theta^T\mu-c)V_x+\frac12\theta^TA\theta\,V_{xx}+U_{\mathrm{consumption}}(c)\right\}=0,\qquad V(x,T)=U_{\mathrm{wealth}}(x).}
$$
The controls are progressively measurable, the investment integrals exist, consumption is nonnegative and locally integrable, and the resulting <portfolio wealth> stays in the utility domain. At zero consumption use the right limit of the <utility function> if needed. This is the usual <admissible trading strategy> formulation; without restrictions preserving the domain the reward need not be defined.
For the verification bound, assume $V$ is a nonnegative classical solution, continuous through the terminal time, with one time <derivative> and two wealth <derivatives> in the interior. Fix any admissible controls and put
$$
R_t=V(X_t,t)+\int_0^tU_{\mathrm{consumption}}(C_s)\,ds.
$$
The <Itô formula> gives
$$
dR_t=\left[V_t+(\theta_t^T\mu-C_t)V_x+\frac12\theta_t^TA\theta_tV_{xx}+U_{\mathrm{consumption}}(C_t)\right]dt+V_x\theta_t^T\sigma\,dW_t.
$$
The drift is nonpositive because the actual controls cannot exceed the supremum in the <Hamilton-Jacobi-Bellman equation>. Localize by <stopping times> on which the <stochastic integral> is a true <martingale>, the derivatives and wealth stay in their interior domain, and the accumulated reward is finite. For each such stop $\tau_n$,
$$
\mathbb E R_{T\wedge\tau_n}\leq V(X_0,0).
$$
Both utilities and $V$ are nonnegative. Thus the <Fatou lemma>, continuity, and the terminal condition give
$$
\boxed{\mathbb E\left[U_{\mathrm{wealth}}(X_T)+\int_0^T U_{\mathrm{consumption}}(C_s)\,ds\right]\leq V(X_0,0).}
$$
For admissible paths strictly positive on the closed horizon, the domain-localizing stops eventually reach $T$. If zero wealth is allowed, the same proof uses the continuous boundary extension of $V$ and the utilities there. Nonnegativity is what permits removing the stops without assuming the uncontrolled stochastic term already has zero expectation. This is <verification by a nonnegative control supermartingale>.
The square-root case needs an additional no-<arbitrage> qualification because the printed volatility matrix need not be invertible. If there is $v\in\ker\sigma^T$ with $v^T\mu>0$, investment $\theta=Kv$ has zero volatility and arbitrarily large positive drift. With any fixed positive consumption, sufficiently large $K$ preserves positive wealth and makes terminal square-root utility tend to infinity. For example, in one dimension $\sigma=0$, $\mu=1$, $\theta=K$, $C=1$ gives $X_s=x+(K-1)(s-t)$, so the value is infinite when $T>t$. \b[A finite separated solution does not exist for all matrices allowed by the printed hypotheses.]
The needed condition is $\mu\in\operatorname{Range}A$, equivalent to $\mu$ being orthogonal to $\ker\sigma^T$, because $A$ is a <symmetric matrix> and $\ker A=\ker\sigma^T$. Assume this condition and let $B=A^+$ be its <Moore-Penrose inverse>; if $A$ is invertible, simply use $B=A^{-1}$. Put $q=\mu^TB\mu\geq0$. Seek the <square-root investment-consumption value> in the form
$$
V(x,t)=2\sqrt{x}\,g(t),\qquad g(T)=1,\qquad g(t)>0.
$$
Then $V_x=g/\sqrt{x}$ and $V_{xx}=-g/(2x^{3/2})$. Completing the quadratic investment expression, using $AB\mu=\mu$, shows that its maximum is attained at $\theta^*=2xB\mu$ and has value $qg\sqrt{x}$:
$$
\frac{g}{\sqrt{x}}\theta^T\mu-\frac{g}{4x^{3/2}}\theta^TA\theta=qg\sqrt{x}-\frac{g}{4x^{3/2}}(\theta-2xB\mu)^TA(\theta-2xB\mu).
$$
Any investment component in $\ker A$ has no effect. For consumption, set $z=\sqrt c$. Its contribution is the concave quadratic $2z-(g/\sqrt{x})z^2$, whose maximum is $\sqrt{x}/g$ at $c^*=x/g^2$. Substituting both maxima into the <Hamilton-Jacobi-Bellman equation> gives
$$
2g'+qg+\frac1g=0.
$$
With $h=g^2$, this becomes $h'+qh+1=0$, $h(T)=1$. Solving this linear <ordinary differential equation> gives
$$
\boxed{h(t)=\begin{cases}e^{q(T-t)}+\dfrac{e^{q(T-t)}-1}{q},&q>0,\\1+T-t,&q=0,\end{cases}\qquad V(x,t)=2\sqrt{x}\sqrt{h(t)}.}
$$
In particular $h\geq1$, so $g=\sqrt h$ stays positive, and the two controls are
$$
\boxed{\theta_t^*=2X_tB\mu,\qquad C_t^*=\frac{X_t}{h(t)}.}
$$
These controls are admissible and actually attain the value. Indeed their <portfolio wealth> satisfies
$$
\frac{dX_s}{X_s}=\left(2q-\frac1{h(s)}\right)ds+2\mu^TB\sigma\,dW_s.
$$
Since $\lVert2\mu^TB\sigma\rVert^2=4q$, the explicit positive solution starting at $(x,t)$ is
$$
X_s=x\exp\left[-\int_t^s\frac{dr}{h(r)}+2\mu^TB\sigma(W_s-W_t)\right].
$$
Its finite-horizon moments of every positive order are finite. More directly,
$$
\mathbb E\sqrt{X_s}=\sqrt{x}\exp\left[\frac q2(s-t)-\frac12\int_t^s\frac{dr}{h(r)}\right].
$$
Differentiating $g(s)\mathbb E\sqrt{X_s}$ and using the equation for $g$ gives $\frac{d}{ds}(g(s)\mathbb E\sqrt{X_s})=-\mathbb E\sqrt{X_s}/g(s)$. Integrating from $t$ to $T$ therefore yields
$$
\mathbb E\left[2\sqrt{X_T}+\int_t^T2\sqrt{C_s^*}\,ds\right]=2g(t)\sqrt{x}.
$$
This proves attainment without relying on an unjustified expectation of a local <martingale>. When $\mu$ is outside $\operatorname{Range}A$, the <singular-volatility drift arbitrage> counterexample applies instead.
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