= Solution
For a normalized trial state $\psi$, expand it in an orthonormal basis of <energy eigenstates>, $\psi=\sum_nc_n\psi_n$. The <Rayleigh quotient> is then
$$
\langle H\rangle_\psi
=\sum_n|c_n|^2E_n\geq E_0\sum_n|c_n|^2=E_0.
$$
This is the <quantum variational principle>: minimizing the expectation of the <Hamiltonian operator> over any trial family gives an upper bound on the <ground-state energy>.
Let $\psi$ be the normalized ground-state wavefunction and form the normalized dilation $\psi_\lambda(x)=\lambda^{1/2}\psi(\lambda x)$. Its kinetic and potential expectations are
$$
T(\lambda)=\lambda^2T,
\qquad
V(\lambda)=\lambda^{-n}V.
$$
The true ground state makes the energy stationary at $\lambda=1$, so
$$
0=\left.\frac d{d\lambda}\bigl(\lambda^2T+\lambda^{-n}V\bigr)\right|_{\lambda=1}
=2T-nV.
$$
Thus the <power-law quantum virial theorem> gives
$$
\boxed{2\langle T\rangle_0=n\langle V\rangle_0.}
$$
If $n\leq-3$ and $A>0$, the two sides have opposite signs, because $T>0$ and $V>0$. If $A<0$, the scaled energy $\lambda^2T+\lambda^{-n}V$ tends to $-\infty$ as $\lambda\to\infty$, since $-n>2$, so the Hamiltonian is not bounded below. If $A=0$, the <free particle> has no normalizable ground state. Hence there is no normalizable ground state for $n\leq-3$.
For $n=1$ take the normalized <Gaussian function>
$$
\psi_\alpha(x)=\left(\frac{2\alpha^2}{\pi}\right)^{1/4}e^{-\alpha^2x^2}.
$$
The <Gaussian integral> gives
$$
\langle T\rangle_\alpha=\frac{\hbar^2\alpha^2}{2m},
\qquad
\langle |x|\rangle_\alpha=\frac1{\sqrt{2\pi}\alpha},
$$
and therefore
$$
E(\alpha)=\frac{\hbar^2\alpha^2}{2m}
+\frac{A}{\sqrt{2\pi}\alpha}.
$$
Its unique minimum satisfies
$$
\alpha^3=\frac{mA}{\sqrt{2\pi}\hbar^2}.
$$
At this point $\langle V\rangle=2\langle T\rangle$, consistently with the virial theorem, and the resulting variational estimate is
$$
\boxed{E_0\leq
\frac32\left(\frac{A^2\hbar^2}{2\pi m}\right)^{1/3}.}
$$
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