= Solution
Since
$$
-V_0<V(x)\leq0
$$
except at the single point $x=0$, and the kinetic-energy expectation is nonnegative, every normalized wavefunction satisfies
$$
\langle H\rangle>-V_0.
$$
Therefore
$$
E_0>-V_0.
$$
For the upper bound, use the admissible choice
$$
a=V_0^2.
$$
Then
$$
\begin{aligned}
E(V_0^2)
&=\frac12V_0^2
-\frac{V_0^2}{\sqrt{1+V_0^2}}\\
&=-\frac12V_0^2+\frac12V_0^4+O(V_0^6).
\end{aligned}
$$
The <Rayleigh-Ritz variational principle> now gives
$$
\boxed{
-V_0<E_0
\leq-\frac12V_0^2+O(V_0^3)}.
$$
Indeed, the displayed calculation provides the sharper remainder $O(V_0^4)$.
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