Step-by-step proof of the First Hohenberg-Kohn Theorem in DFT

I’m trying to proof the first Hohenberg-Kohn theorem using the Reductio ad absurdum, but I’m stucked in the last point. First, I know that the eletronic density is:
$${ρ(r)=∑_{i=1}^n|χ_i(r)|²}$$
And,
$$Ψ = ∑_{i=1}^nc_iχ_i(r)$$
The First Theorem states that the external potential $v(r)$ is determined, within a trivial additive constant, by the electron density $ρ(r)$. To proof this, let A and B two systems having the same electron density $ρ(r)$ but two different external potentials, so,
$$⟨Ψ_A|H_A|Ψ_B⟩=E_A$$ $$⟨Ψ_B|H_B|Ψ_B⟩=E_B$$
Using the variational principle, and using ${Ψ_A}$ as a trial function for the system B, and ${Ψ_B}$ as a trial function for the system A, we have:
$$⟨Ψ_B|H_A|Ψ_B⟩>E_A rightarrow ⟨Ψ_B|H_B|Ψ_B⟩-⟨Ψ_B|H_B|Ψ_B⟩+⟨Ψ_B|H_A|Ψ_B⟩>E_A rightarrow boldsymbol{E_B +⟨Ψ_B|H_A-H_B|Ψ_B⟩ > E_A} $$
$$⟨Ψ_A|H_B|Ψ_A⟩>E_B rightarrow ⟨Ψ_A|H_A|Ψ_A⟩-⟨Ψ_A|H_A|Ψ_A⟩+⟨Ψ_A|H_B|Ψ_A⟩>E_B rightarrow boldsymbol{E_A -⟨Ψ_A|H_A-H_B|Ψ_A⟩ > E_B} $$
The two Hamiltonians only differ on the potential, so,
$$boldsymbol{E_B +⟨Ψ_B|V_A-V_B|Ψ_B⟩ > E_A} $$
$$boldsymbol{E_A -⟨Ψ_A|V_A-V_B|Ψ_A⟩ > E_B} $$
All textbooks and online materials I’ve read, transform the $⟨Ψ_A|V_A-V_B|Ψ_A⟩$ and $⟨Ψ_B|V_A-V_B|Ψ_B⟩$, in the term $∫(V_A-V_B)ρ(r)dr$, but they didn’t show how they do that. So, my question is: How can I do that? I’ve tried this:
$$⟨Ψ_A|V_A-V_B|Ψ_A⟩ = ∫Ψ_A^*(V_A-V_B)Ψ_Adr = ∫(V_A-V_B)∑_{i=1}^nc_i^*χ_i∑_{i=1}^nc_iχ_idr$$
$$⟨Ψ_A|V_A-V_B|Ψ_A⟩ = ∫(V_A-V_B)∑_{i=1}^n|c_i|^2|χ_i|^2dr$$
How do you deal with the coefficients $|c_i|^2$ to substitute for $ρ(r)$? Or the $ρ(r)$ already has this coefficients and am I writing $ρ(r)$ in a wrong way?