= Solution
Differentiate the <Friedmann equation>
$$
H^2=\frac{8\pi G}{3c^2}\rho-\frac{kc^2}{R^2a^2}
$$
and use the <cosmological perfect-fluid continuity equation>
$$
\dot\rho=-3H(\rho+P).
$$
After dividing by $2H$,
$$
\dot H=-\frac{4\pi G}{c^2}(\rho+P)
+\frac{kc^2}{R^2a^2}.
$$
Since $\ddot a/a=\dot H+H^2$, eliminating the curvature term gives the <Friedmann acceleration equation>
$$
\boxed{
\frac{\ddot a}{a}
=-\frac{4\pi G}{3c^2}(\rho+3P)
}.
$$
Under the <strong energy condition in a Friedmann universe>, $\rho+3P\geq0$, so $\ddot a/a\leq0$. Therefore
$$
\dot H=\frac{\ddot a}{a}-H^2\leq-H^2,
$$
and wherever $H\ne0$,
$$
\boxed{
\frac d{dt}H^{-1}
=-\frac{\dot H}{H^2}\geq1
}.
$$
If $H(t_0)>0$, integrating the inequality backward shows that $H^{-1}$ reaches zero no later than $t_0-H(t_0)^{-1}$. Thus $H\to+\infty$ and $a\to0$ at a finite time in the past. If $H(t_0)<0$, the same argument forward in time makes $H\to-\infty$ and $a\to0$ at a finite future time. The sign of $H$ distinguishes an expanding universe with a past Big Bang from a contracting universe ending in a <Big Crunch>. Hence a Friedmann universe obeying the strong energy condition cannot remain nonsingular for unlimited proper time in both directions.
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