= Solution
Expand the trial <wavefunction> in a complete <orthonormal set> of exact <eigenfunctions>, $\psi=\sum_n c_n\phi_n$, with $\hat H\phi_n=E_n\phi_n$ and $E_n\ge E_0$. The coefficients may depend on any chosen variational parameters. <Orthogonality> removes the cross terms in both integrals, giving
$$
\langle\psi|\psi\rangle=\sum_n|c_n|^2,\qquad
\langle\psi|\hat H|\psi\rangle=\sum_n E_n|c_n|^2.
$$
Consequently
$$
\boxed{\mathcal E[\psi]-E_0=\frac{\sum_n(E_n-E_0)|c_n|^2}{\sum_n|c_n|^2}\ge0.}
$$
This proves the <quantum variational principle>. The denominator is positive because the trial function is nonzero. Equality requires every occupied exact state to have $E_n=E_0$, allowing for <ground-state degeneracy>. With a continuous part of the spectrum, the corresponding nonnegative spectral integral gives the same conclusion.
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