= Solution
Give $H_0^1$ the equivalent <Hilbert space> norm $\|u\|_H=\|\nabla u\|_2$, using the <Poincaré inequality>. Define
$$
a(u,v)=\int_\Omega\nabla u\cdot\nabla v+\int_\Omega(b\cdot\nabla u)v,\qquad\ell(v)=\int_\Omega fv.
$$
The <Cauchy-Schwarz inequality> gives $|a(u,v)|\leq(1+C_P\|b\|_\infty)\|u\|_H\|v\|_H$ and $|\ell(v)|\leq C_P\|f\|_2\|v\|_H$.
For $\varphi\in C_0^\infty(\Omega)$, the definition of <weak divergence>, applied to $\varphi^2/2$, gives
$$
\int_\Omega(b\cdot\nabla\varphi)\varphi=-\frac12\int_\Omega(\operatorname{div}b)\varphi^2\geq0.
$$
The left side is continuous in the $H_0^1$ norm. Approximate any $u\in H_0^1$ by <test functions> and pass to the limit on the left to obtain $\int(b\cdot\nabla u)u\geq0$. This density argument is essential: local integrability of $\operatorname{div}b$ alone does not justify treating $(\operatorname{div}b)u^2$ as integrable for arbitrary $u\in H_0^1$.
Consequently $a(u,u)\geq\|u\|_H^2$, so $a$ is a <coercive bilinear form>. The <Lax-Milgram theorem> supplies a unique <weak solution> with $\boxed{\|\nabla u\|_2\leq C_P\|f\|_2}$. The reusable estimate is <coercivity from a drift with nonpositive weak divergence>.
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