= Solution
For a nonzero normalized or unnormalized trial function in the Hamiltonian's quadratic-form domain, the <Rayleigh-Ritz variational principle> gives $E_0\leq\langle\psi,H\psi\rangle/\langle\psi,\psi\rangle$. Expanding in energy eigenstates, or using the spectral measure, makes the quotient a weighted average of spectral energies, all at least $E_0$. For an even one-dimensional potential, the <ground state> is even and the first excited <bound state> is odd by the nodal ordering theorem. Restricting to odd trial functions therefore gives an upper bound on the lowest odd energy and on $E_1$ when that <bound state> exists; a negative odd quotient also certifies its existence.
Here $\hbar^2/(2m)=1$. For the Gaussian trial function, <integration by parts> gives kinetic quotient $\int|\psi'|^2/\int|\psi|^2=a/2$, and the potential quotient is $-V_0\sqrt{a/(1+a)}$. Therefore
$$
\boxed{E_0\leq E(a)=\frac a2-V_0\sqrt{\frac a{1+a}}.}
$$
For $a>0$, its zero obeys $a(1+a)=4V_0^2$, giving the unique positive zero
$$
\boxed{a_z=\frac{\sqrt{1+16V_0^2}-1}{2}.}
$$
Also $E'(a)=1/2-V_0/[2\sqrt a(1+a)^{3/2}]$, so its stationary points satisfy $a(1+a)^3=V_0^2$. The left side increases strictly from zero to infinity, proving that there is exactly one stationary point, a minimum.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ii/paper-1-gaussian-variational-bound.png]
{title=Gaussian variational energy with its negative minimum and positive zero}
As $a\downarrow0$, $E(a)=-V_0\sqrt a+O(a)$ approaches zero from below; as $a\to\infty$, it tends to infinity. The sketch and variational inequality therefore give a negative energy for every $V_0>0$. Since this decaying potential has <essential spectrum> beginning at zero, \b[at least one <bound state> exists for every positive well depth].
For small $V_0$, the minimizer is $a_*=V_0^2-3V_0^4+O(V_0^6)$. Substitution gives $E(a_*)=-V_0^2/2+V_0^4/2+O(V_0^6)$. Meanwhile the potential is strictly greater than $-V_0$ except at a single point and the <kinetic energy> is nonnegative, so a normalized bound-state expectation is strictly greater than $-V_0$. Thus
$$
\boxed{-V_0<E_0\leq\epsilon(V_0),\qquad
\epsilon(V_0)=-\frac12V_0^2+O(V_0^4).}
$$
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