= Solution
Use the normalized scaled trial state
$$
\psi_\lambda(x)=\lambda^{1/2}\psi_0(\lambda x),
\qquad \lambda>0.
$$
If
$$
T_0=\langle\psi_0,T\psi_0\rangle>0,
\qquad
V_0=\langle\psi_0,V\psi_0\rangle<0,
$$
then scaling derivatives and changing variables give
$$
E(\lambda)=\langle\psi_\lambda,H\psi_\lambda\rangle
=\lambda^2T_0+\lambda^nV_0.
$$
If a finite ground state existed, $\lambda=1$ would minimize this expression, and hence
$$
0=E'(1)=2T_0+nV_0.
$$
It would follow that
$$
E_0=T_0+V_0=\left(1-\frac2n\right)T_0>0
$$
for $n>2$. But a bound state of an attractive potential tending to zero at infinity must have negative energy. This is a contradiction.
Indeed, the scaling gives the stronger collapse statement
$$
E(\lambda)=\lambda^2T_0+\lambda^nV_0\longrightarrow-\infty
\qquad(\lambda\to\infty)
$$
when $n>2$. Thus the ideal singular Hamiltonian is not bounded below and has no finite ground bound state.
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