= Solution
The <Hubble parameter> is the fractional expansion rate
$$
H=\frac{\dot a}{a}.
$$
Here $\rho$ is the homogeneous energy density, $k\in\{-1,0,1\}$ records the sign of the <spatial curvature of an FLRW universe>, and $R$ is the fixed curvature length scale used with the dimensionless scale factor $a(t)$.
Energy-momentum conservation in a homogeneous expanding universe gives the continuity equation
$$
\dot\rho+3H(\rho+P)=0.
$$
Differentiate the <Friedmann equation>
$$
H^2=\frac{8\pi G}{3c^2}
\left(\rho-\frac{kc^2}{R^2a^2}\right).
$$
Using $\dot a/a=H$ and the continuity equation gives
$$
\dot H
=-\frac{4\pi G}{c^2}(\rho+P)
+\frac{8\pi G}{3c^2}\frac{kc^2}{R^2a^2}.
$$
Eliminating the curvature term with the Friedmann equation yields
$$
\dot H+H^2
=-\frac{4\pi G}{3c^2}(\rho+3P).
$$
Since $\dot H+H^2=\ddot a/a$, this is the <Raychaudhuri equation>
$$
\boxed{\frac{\ddot a}{a}
=-\frac{4\pi G}{3c^2}(\rho+3P)}.
$$
Under the <strong energy condition> $\rho+3P\ge0$, one has $\ddot a\le0$. Thus $a(t)$ is concave, and for $t<t_0$ it lies below its tangent at the present time:
$$
a(t)\le a(t_0)+\dot a(t_0)(t-t_0)
=a(t_0)\bigl[1-H_0(t_0-t)\bigr].
$$
The right side vanishes at $t=t_0-H_0^{-1}$. A positive scale factor therefore cannot extend regularly to that time; it must reach the <Big Bang> singularity at some $t_{\rm BB}\ge t_0-H_0^{-1}$. Equivalently,
$$
\boxed{t_0-t_{\rm BB}\le H_0^{-1}}.
$$
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