= Solution
Use the paper's positive binding-energy convention $U>0$; the physical <Newtonian gravitational potential energy> is $-U$. Assume positive <masses>, an isolated system, and no collision during the time interval under consideration. Differentiate half the scalar <moment of inertia>, $I$, twice:
$$
\dot I=\sum_km_k\boldsymbol r_k\cdot\boldsymbol v_k,
\qquad
\ddot I=\sum_km_kv_k^2+\sum_k\boldsymbol r_k\cdot\boldsymbol F_k.
$$
Group the <force> term into unordered pairs. The <force> on particle $j$ from $k$ is $-Gm_jm_k(\boldsymbol r_j-\boldsymbol r_k)/r_{jk}^3$; the two contributions to the virial sum are therefore
$$
\boldsymbol r_j\cdot\boldsymbol F_{jk}
+\boldsymbol r_k\cdot\boldsymbol F_{kj}
=-\frac{Gm_jm_k}{r_{jk}}.
$$
Adding them proves the instantaneous <virial theorem> identity
$$
\boxed{\ddot I=2T-U=T+E,\qquad E=T-U.}
$$
No time average is needed for this form. A vanishing average of $\ddot I$ would require additional boundedness assumptions.
For the <pairwise moment-of-inertia identity>, expand the squared pair distances:
$$
\sum_{j<k}m_jm_k|\boldsymbol r_j-\boldsymbol r_k|^2
=M\sum_km_k r_k^2-\left|\sum_km_k\boldsymbol r_k\right|^2.
$$
The last term vanishes in the <centre of mass> frame, giving
$$
\boxed{I=\frac1{2M}\sum_{j<k}m_jm_k r_{jk}^2.}
$$
Let $C=\sum_{j<k}m_jm_k$, and label the two smallest <masses> $m_{(1)},m_{(2)}$. If $r=\min r_{jk}$, every term in $U$ is at most $Gm_jm_k/r$, so $U\leq GC/r$. One pair attains $r$, and its <mass> product is at least $m_{(1)}m_{(2)}$, so $U\geq Gm_{(1)}m_{(2)}/r$. Hence
$$
\boxed{B_0\leq rU\leq A_0,\qquad
B_0=Gm_{(1)}m_{(2)},\quad A_0=GC.}
$$
These constants show the <minimum-separation binding-energy bounds>: inverse binding <energy> is comparable to the nearest-pair distance, independently of configuration.
Similarly, every separation is at most $R=\max r_{jk}$ and at least one pair attains $R$. The pairwise identity gives the <maximum-separation inertia bounds>
$$
\boxed{B_1R^2\leq I\leq A_1R^2,\qquad
B_1=\frac{m_{(1)}m_{(2)}}{2M},\quad A_1=\frac C{2M}.}
$$
Both lower bounds rely on actual extremal separations, rather than an arbitrary numerical lower or upper estimate for all the pair distances.
\b[Negative <energy> does not imply a positive lower bound on the minimum separation.] The printed request is false if it is meant to exclude close approaches or collisions; the trivial bound $r\geq0$ says nothing of that kind. In fact $E<0$ and $T\geq0$ give $U=T-E\geq|E|$, which, combined with $rU\leq A_0$, proves the <negative-energy minimum-separation upper bound>
$$
\boxed{r\leq\frac{A_0}{|E|}.}
$$
This ensures at least one close pair, not confinement of every particle or prevention of collision.
An explicit <negative-energy gravitational collision> is a pair initially at rest with separation $a$. Its total <energy> is $E=-Gm_1m_2/a<0$. For <reduced mass> $\mu=m_1m_2/(m_1+m_2)$, its relative radial equation is
$$
\frac12\mu\dot r^2-\frac{Gm_1m_2}{r}=-\frac{Gm_1m_2}{a},\qquad
\dot r=-\sqrt{2G(m_1+m_2)\left(\frac1r-\frac1a\right)}.
$$
The separation decreases to zero in the finite time
$$
t_{\mathrm{coll}}=\int_0^a\frac{dr}{\sqrt{2G(m_1+m_2)(1/r-1/a)}}
=\pi\sqrt{\frac{a^3}{8G(m_1+m_2)}}.
$$
Before that time the motion is a regular Newtonian solution, yet its separation has no positive infimum. Even at fixed negative <energy>, bound <Kepler orbits> with eccentricity approaching one have arbitrarily small <pericentre distance>. Thus a missing <angular momentum> or collision-exclusion hypothesis cannot be supplied by the <energy> sign alone.
For $E>0$, the instantaneous <virial theorem> gives $\ddot I=T+E\geq E$. Integrating twice from any regular time $t_0$ gives
$$
I(t)\geq I(t_0)+\dot I(t_0)(t-t_0)+\frac E2(t-t_0)^2.
$$
Using $I\leq A_1R^2$, obtain the <positive-energy linear diameter growth> bound
$$
\boxed{R(t)\geq
\sqrt{\frac{I(t_0)+\dot I(t_0)(t-t_0)+E(t-t_0)^2/2}{A_1}}}
$$
whenever the numerator is nonnegative. Consequently, for a solution existing for arbitrarily large future times,
$$
\liminf_{t\to\infty}\frac{R(t)}{t-t_0}\geq\sqrt{\frac E{2A_1}}>0.
$$
This is the precise at-least-linear expansion statement. It does not assert that $R$ is monotone at every instant, nor that all individual particles escape. The large-time conclusion presupposes continued existence of the trajectory; the inequality itself holds on every nonsingular time interval.
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