The trial wave function can be expanded in the eigenstates \( \Psi_i(\boldsymbol{R}) \) of the hamiltonian since they form a complete set, viz.,
$$ \Psi_T(\boldsymbol{R};\boldsymbol{\alpha})=\sum_i a_i\Psi_i(\boldsymbol{R}), $$and assuming that the set of eigenfunctions are normalized, one obtains
$$ \frac{\sum_{nm}a^*_ma_n \int d\boldsymbol{R}\Psi^{\ast}_m(\boldsymbol{R})H(\boldsymbol{R})\Psi_n(\boldsymbol{R})} {\sum_{nm}a^*_ma_n \int d\boldsymbol{R}\Psi^{\ast}_m(\boldsymbol{R})\Psi_n(\boldsymbol{R})} =\frac{\sum_{n}a^2_n E_n} {\sum_{n}a^2_n} \ge E_0, $$where we used that \( H(\boldsymbol{R})\Psi_n(\boldsymbol{R})=E_n\Psi_n(\boldsymbol{R}) \).