= Solution
Use <geometrized units> $G=c=1$ and <metric signature> $(-+++)$; $\tau$ is <proper time>. The <Killing vectors> $\partial_t$ and $\partial_\phi$ of the <Schwarzschild metric> give the conserved specific <Killing energy> and specific <angular momentum>
$$
e=\frac Em=\left(1-\frac{2M}{r}\right)\dot t=1,\qquad
h=r^2\dot\phi.
$$
Put $f=1-2M/r$. In the equatorial plane $\dot\theta=0$, and normalization of the <four-velocity> gives
$$
-1=-f\dot t^2+f^{-1}\dot r^2+r^2\dot\phi^2,
\qquad
\dot r^2=1-f\left(1+\frac{h^2}{r^2}\right)
=\frac{2Mr^2-h^2(r-2M)}{r^3}.
$$
Choosing the inward branch therefore yields
$$
\boxed{u^\mu=\left(\frac r{r-2M},\
-\frac{\sqrt{2Mr^2-h^2(r-2M)}}{r^{3/2}},\
0,\ \frac h{r^2}\right).}
$$
The divergence of the time component at the <Schwarzschild event horizon> is a coordinate effect; the inward radial component tends to $-1$.
\b[Capture from infinity.] To reach the <event horizon> from infinity, the radial square must remain nonnegative throughout $r>2M$. Equivalently,
$$
h^2\leq\frac{2Mr^2}{r-2M}\quad\text{for all }r>2M.
$$
Differentiating the right side gives $2Mr(r-4M)/(r-2M)^2$, so its minimum occurs at $r=4M$ and equals $16M^2$. Thus
$$
\boxed{|h|\leq4M}
$$
is the necessary capture bound. At equality the radial numerator is $2M(r-4M)^2$. An inward particle arriving from larger radii approaches the unstable orbit $r=4M$ only after infinite <proper time>, because $\dot r$ is proportional to $-(r-4M)$ near that orbit. Actual plunges from infinity require $|h|<4M$. This is <Schwarzschild marginally bound capture>.
The origin-at-infinity hypothesis is important and is not explicit in the PDF. A particle already inside the angular-momentum barrier can plunge with larger $|h|$. For example, $h=10M$ and initial radius $r_0=2.01M$ give positive radial numerator $7.0802M^3$, remaining positive as $r$ decreases to $2M$. This trajectory has $e=1$ and reaches the <event horizon>. Thus an unrestricted claim about every inward particle would be false; the bound is the intended capture-from-infinity statement.
\b[Invariant collision energy.] At a collision, the total <four-momentum> is $P^\mu=m(u_1^\mu+u_2^\mu)$. The invariant <center-of-mass energy> uses the covariant metric:
$$
E_{\rm com}^2=-g_{\mu\nu}P^\mu P^\nu.
$$
The PDF instead prints a raised metric multiplying raised velocities. That contraction is not a tensor scalar; the corrected expression above, or a raised metric with lowered momenta, is required. Since each <four-velocity> has norm $-1$,
$$
\boxed{E_{\rm com}^2=2m^2(1-g_{\mu\nu}u_1^\mu u_2^\nu).}
$$
Let $D_i=2Mr^2-h_i^2(r-2M)$. For two inward trajectories the radial product is positive, and direct substitution into the <Schwarzschild metric> gives
$$
g_{\mu\nu}u_1^\mu u_2^\nu
=-\frac r{r-2M}
+\frac{\sqrt{D_1D_2}}{r^2(r-2M)}
+\frac{h_1h_2}{r^2}.
$$
Putting these terms over one denominator proves
$$
\boxed{E_{\rm com}^2=
\frac{2m^2}{r^2(r-2M)}
\left[2r^2(r-M)-h_1h_2(r-2M)-\sqrt{D_1D_2}\right].}
$$
\b[Horizon limit and the upper bound.] A cancellation-free way to take the limit is to write $a_i=1+h_i^2/r^2$ and $\dot r_i=-\sqrt{1-fa_i}$. As $f\to0^+$,
$$
\sqrt{(1-fa_1)(1-fa_2)}
=1-\frac f2(a_1+a_2)+O(f^2).
$$
Hence
$$
1-g(u_1,u_2)
=1+\frac{1-\sqrt{(1-fa_1)(1-fa_2)}}f-\frac{h_1h_2}{r^2}
=2+\frac{(h_1-h_2)^2}{2r^2}+O(f),
$$
and therefore
$$
\boxed{\lim_{r\downarrow2M}E_{\rm com}^2
=m^2\left[4+\frac{(h_1-h_2)^2}{4M^2}\right].}
$$
For particles captured from infinity, $|h_1-h_2|\leq8M$, giving \b[$E_{\rm com}\leq m\sqrt{20}$] in the horizon limit. For actual captured trajectories the inequality is strict, but the supremum is approached by $h_1\to4M^-$ and $h_2\to-4M^+$. Their azimuthal starting positions can be chosen so that the trajectories meet. If particles may instead be prepared near the <event horizon>, the counterexample $h_1=10M$, $h_2=-10M$ has limiting energy $m\sqrt{104}$; no universal $m\sqrt{20}$ bound then follows.
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