= Solution
It is enough to choose an admissible parameter, without solving the minimizing equation. Take $a=V_0^2>0$ in the <Gaussian variational bound for an attractive Gaussian well>. The <Rayleigh-Ritz variational principle> gives
$$
E_0\leq\frac{V_0^2}{2}-\frac{V_0^2}{\sqrt{1+V_0^2}}=-\frac{V_0^2}{2}+\varepsilon,
$$
where an explicit remainder is
$$
\varepsilon=V_0^2\left(1-\frac1{\sqrt{1+V_0^2}}\right)=\frac{V_0^4}{\sqrt{1+V_0^2}(\sqrt{1+V_0^2}+1)}.
$$
The denominator tends to $2$ as $V_0\to0$, proving $\varepsilon=O(V_0^4)$ in <Big O notation>. More precisely the <binomial series> gives
$$
\varepsilon=\frac12V_0^4-\frac38V_0^6+O(V_0^8),
$$
and therefore
$$
\boxed{E_0\leq-\frac12V_0^2+\varepsilon,\qquad\varepsilon=O(V_0^4).}
$$
This is an upper bound on the exact <ground state> energy, rather than an assertion that the leading coefficient of its exact weak-potential expansion is $-1/2$.
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