= Solution
For a <self-adjoint> <Hamiltonian operator> bounded below with <ground-state energy> $E_0$, every admissible nonzero trial <wavefunction> has a <Rayleigh quotient> at least $E_0$. To see this explicitly in a complete <orthonormal eigenbasis>, write $|\psi\rangle=\sum_n c_n|n\rangle$ with $H|n\rangle=E_n|n\rangle$. Then
$$
\frac{\langle\psi|H|\psi\rangle}{\langle\psi|\psi\rangle}-E_0
=\frac{\sum_n|c_n|^2(E_n-E_0)}{\sum_n|c_n|^2}\geq0,
$$
so
$$
\boxed{E_0\leq\frac{\langle\psi|H|\psi\rangle}{\langle\psi|\psi\rangle}.}
$$
For continuous spectral components, the <spectral theorem> replaces the sum by an integral with the same nonnegative integrand. Equality requires the trial <quantum state> to lie entirely in the <ground-state subspace>. The <variational method> chooses a tractable family, evaluates the quotient and minimizes over its parameters. This gives an upper bound; enlarging the family can only improve that bound. The trial must belong to the <operator domain> for the operator expression above, or have finite energy in the corresponding quadratic-form version of the <quantum variational principle>.
For the oscillator, use the standard kinetic operator $T=-\hbar^2d^2/(2m\,dx^2)$: the original PDF omits the superscript $2$ on the derivative in its numerator. Take $a>0$, and normalize the scaled exact <ground state> as
$$
\psi_\lambda(x)=\sqrt\lambda\,\phi_0(\lambda x),\qquad\lambda>0.
$$
The factor $\sqrt\lambda$ is optional in a quotient but makes the calculation transparent. Changing variable to $u=\lambda x$ and differentiating the scaled <wavefunction> gives
$$
\langle T\rangle_\lambda=\lambda^2T_0,\qquad
\langle V\rangle_\lambda=\lambda^{-2}V_0,
\qquad T_0=\langle\phi_0|T|\phi_0\rangle,\quad V_0=\langle\phi_0|ax^2|\phi_0\rangle.
$$
At $\lambda=1$ this trial is the exact minimizing state, so its energy $\mathcal E(\lambda)=\lambda^2T_0+\lambda^{-2}V_0$ has zero derivative there. Hence $2T_0-2V_0=0$. Since $E_0=T_0+V_0$,
$$
\boxed{\langle T\rangle=\langle V\rangle=E_0/2.}
$$
This is the <quantum virial identity from scaling> for a quadratic potential, and hence the oscillator form of the <virial theorem>.
For the inverse-power attraction, take $a>0$ and suppose, for contradiction, that an exact <ground state> exists with separately finite <kinetic energy> $T_0>0$ and potential mean $U_0<0$, admitting the same dilation variations. Homogeneity now gives
$$
\mathcal E(\lambda)=\lambda^2T_0+\lambda^pU_0.
$$
Stationarity at $\lambda=1$ would require $2T_0+pU_0=0$. Its second derivative would then be
$$
\mathcal E''(1)=2T_0+p(p-1)U_0=2(2-p)T_0.
$$
For $p>2$ this is negative, contradicting a ground-state minimum. More strongly, one does not need a presumed exact state: choose any normalized smooth test function supported away from the singular origin. Its $T$ and $U$ are finite, and its scaled quotient satisfies
$$
\boxed{\mathcal E(\lambda)=\lambda^2T+\lambda^pU\longrightarrow-\infty
\quad(\lambda\to\infty,\ p>2).}
$$
The unregularized attraction therefore exhibits <fall to the centre for a supercritical inverse-power potential> and has no stable <ground state>.
There is a necessary qualification to the PDF's broader wording about “no bound state”. Under the usual finite-energy <virial theorem> assumptions, a putative negative-energy <bound state> would satisfy
$$
E=T+U=\left(1-\frac2p\right)T>0,
$$
which contradicts being below the continuum threshold at zero. Thus no negative-energy bound state satisfying those regularity assumptions is possible. But this singular differential expression needs a specified domain at $x=0$; separately divergent kinetic and potential means or boundary conditions not preserved by dilation invalidate that ordinary calculation. The <domain assumptions in singular-potential virial arguments> must therefore be stated: the scaling proof establishes absence of a semibounded ground-state problem, rather than a domain-independent theorem about every formal square-integrable solution. A short-distance regularization would be a different potential.
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