= Solution
As $a\downarrow0$,
$$
E(a)=-V_0\sqrt a+O(a),
$$
so $E(a)<0$ for all sufficiently small positive $a$. On the other hand,
$$
E(a)\longrightarrow+\infty
\qquad(a\to\infty).
$$
The derivative satisfies
$$
E'(a)\to-\infty\quad(a\downarrow0),
\qquad
E'(a)\to\frac12\quad(a\to\infty).
$$
The extremum equation
$$
a(1+a)^3=V_0^2
$$
has exactly one positive solution because its left side is strictly increasing. Thus $E$ has one global minimum, and that minimum is negative.
The potential tends to zero at spatial infinity, so zero is the threshold of the continuous spectrum. A trial state with negative energy expectation forces the spectrum below zero; the corresponding lowest spectral value is a normalizable <bound state>. Hence every $V_0>0$ produces at least one bound state.
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