= Solution
Expand the trial in a complete <orthonormal eigenbasis> of the <Quantum Hamiltonian>, $\widetilde\Psi=\sum_nc_n\phi_n$, with $\hat H\phi_n=E_n\phi_n$. <Orthonormality> gives
$$
\widetilde E=
\frac{\sum_n|c_n|^2E_n}{\sum_n|c_n|^2},\qquad
\widetilde E-E_0=
\frac{\sum_n|c_n|^2(E_n-E_0)}{\sum_n|c_n|^2}\geq0.
$$
Every summand in the numerator is nonnegative because $E_n\geq E_0$. This proves the <quantum variational principle> for the stated discrete expansion; the trial <energy> is a <weighted mean> of exact <energy eigenvalues>.
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