= Solution
The function $f$ tends to $-\infty$ both as $r\to0^+$ and as $r\to\infty$. Its only positive stationary point is
$$
r_0=\left(\frac m{H^2}\right)^{1/3},
$$
and it is a maximum with
$$
f(r_0)=1-3(mH)^{2/3}.
$$
Therefore $f$ has two positive simple roots exactly when
$$
mH<\frac1{3\sqrt3}=\frac1{\sqrt{27}}.
$$
At $r_c=1/(\sqrt3H)$,
$$
f(r_c)=\frac23-2\sqrt3mH>0.
$$
Since $f$ is positive precisely between its two positive roots, this proves
$$
\boxed{r_-<\frac1{\sqrt3H}<r_+.}
$$
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