= Solution
For a normalized trial state $|\psi\rangle=\sum_nc_n|n\rangle$ in the form domain of a bounded-below <Hamiltonian>, the <quantum variational principle> gives $\langle H\rangle=\sum_n|c_n|^2E_n\geq E_0$. Thus each trial <expectation> is an upper bound for the ground-state energy. The same spectral argument covers a continuum: for a decaying potential whose essential spectrum begins at $0$, a negative trial <expectation> guarantees a negative-energy <bound state>.
For the stated exponential trial function, normalization follows from $4\alpha^3\int_0^\infty r^2e^{-2\alpha r}\,dr=1$. The kinetic <expectation>, using $\int|\nabla\psi|^2$, is $\hbar^2\alpha^2/(2m)$. The potential <expectation> is $-4A\alpha^3\int_0^\infty re^{-(\mu+2\alpha)r}\,dr$. Therefore
$$
\boxed{E(\alpha)=\frac{\hbar^2\alpha^2}{2m}-\frac{4A\alpha^3}{(\mu+2\alpha)^2}.}
$$
The <exponential variational bound for the Yukawa potential> at $\alpha=\mu/2$ gives $E=(\mu/8)(\hbar^2\mu/m-A)<0$ if $\mu<Am/\hbar^2$. Thus \b[a <bound state> necessarily exists in the stated range].
For $\mu\geq Am/\hbar^2$, this single-exponential family cannot prove binding: the maximum of $4A\alpha/(\mu+2\alpha)^2$ is $A/(2\mu)$, so every $E(\alpha)\geq0$. This does not prove absence of <bound states>.
The equality case can in fact be strengthened. At $\mu=Am/\hbar^2$, use normalized exponentials $\psi_{\mu/2},\psi_\mu$ and write $Q=\hbar^2\mu^2/m$. Their <matrix> elements are
$$
\langle\psi_{\mu/2},H\psi_{\mu/2}\rangle=0,\quad
\langle\psi_{\mu/2},H\psi_\mu\rangle=-\frac{8\sqrt2}{675}Q,\quad
\langle\psi_\mu,H\psi_\mu\rangle=\frac Q{18}.
$$
The Rayleigh numerator for $\psi_{\mu/2}+\epsilon\psi_\mu$ is negative for sufficiently small $\epsilon>0$, and its norm is nonzero. Hence \b[binding also holds at equality, and by <continuity> in a small interval beyond it]. For arbitrary larger $\mu$, these trials alone do not settle existence; a nonnegative variational upper bound cannot establish a nonnegative ground-state energy.
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