= Solution
For a <coercive Neumann energy with a positive reaction term>, let $\nu>0$ be the ellipticity constant and $A_0=\max_{i,j}\|a_{ij}\|_\infty$. <Cauchy-Schwarz inequality>[Cauchy-Schwarz] gives
$$
|B[u,v]|\le nA_0\|\nabla u\|_2\|\nabla v\|_2+\|c\|_\infty\|u\|_2\|v\|_2
\le C\|u\|_{H^1}\|v\|_{H^1}.
$$
Bilinearity follows from the <integral> and weak-derivative <linearity>. Ellipticity and the strictly positive zeroth-order coefficient give
$$
\boxed{B[u,u]\ge\nu\|\nabla u\|_2^2+\mu\|u\|_2^2\ge\kappa\|u\|_{H^1}^2,\quad\kappa=\min(\nu,\mu)>0.}
$$
This controls constants as well as <gradients>, so no zero-mean constraint or <Neumann boundary condition>[Neumann] compatibility condition is required. Symmetry of the coefficients makes $B$ symmetric.
Write $F(v)=\int_\Omega fv$. This is a <bounded linear functional> on $H^1$, with $|F(v)|\le\|f\|_2\|v\|_{H^1}$. Consequently
$$
I(v)\ge\frac\kappa2\|v\|_{H^1}^2-\|f\|_2\|v\|_{H^1}\longrightarrow+\infty
$$
as the <norm> tends to <infinity>. The infimum is finite, and every <minimizing sequence> is <bounded>. The <Hilbert space> $H^1$ is reflexive, so a minimizing <subsequence> converges <weak convergence>[weakly] to some $u\in H^1$.
To verify lower semicontinuity directly, put $w_j=v_j-u$. Boundedness of $B$ makes $B[u,w_j]\to0$ under <weak convergence>, and symmetry gives
$$
B[v_j,v_j]=B[u,u]+2B[u,w_j]+B[w_j,w_j]\ge B[u,u]+o(1).
$$
Also $F(v_j)\to F(u)$. Therefore $I(u)\le\liminf I(v_j)$, so \b[the absolute minimum is attained].
For every $h\in H^1$ the real <function> $t\mapsto I(u+th)$ has a minimum at zero. Its <derivative> there is $B[u,h]-F(h)$, yielding the required weak equation for every $h$. If two <weak solutions> exist, their difference $z$ satisfies $B[z,h]=0$ for all $h$; choose $h=z$ and use coercivity to obtain $z=0$. In fact the weak equation also gives the exact energy identity
$$
\boxed{I(v)-I(u)=\tfrac12B[v-u,v-u]\ge\tfrac\kappa2\|v-u\|_{H^1}^2.}
$$
Thus the <minimizer> and <weak solution> are both unique. The natural <boundary condition> is encoded by testing in the whole $H^1$ space, without needing <pointwise> <conormal flux of a divergence-form operator>[conormal derivatives] for rough coefficients.
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