= Solution
{c}
For $a>0$, the assumed inequality and the <Friedmann acceleration equation> imply $\ddot a\le0$. Thus $a$ is a <concave function>. Let $a_0=a(t_0)>0$ and $v_0=\dot a(t_0)>0$. At every earlier time in the same regular expanding interval, $\dot a(t)\ge v_0$. Integration gives the <concavity bound on past FLRW evolution>:
$$
a_0-a(t)=\int_t^{t_0}\dot a(s)\,ds\ge v_0(t_0-t),
\qquad
\boxed{a(t)\le a_0-v_0(t_0-t).}
$$
The right-hand side reaches zero at $t=t_0-a_0/v_0$. Hence an everywhere regular positive <scale factor> cannot extend arbitrarily far into the past. With the usual implicit continuation assumption that the cosmological solution remains regular while $a>0$, it reaches $a(t_B)=0$ at a finite past boundary satisfying
$$
\boxed{0<t_0-t_B\le\frac{a_0}{v_0}=\frac1{H(t_0)}.}
$$
\b[Past expansion lasts at most the present inverse <Hubble parameter>.] The inequality is the essential argument, and equality is attained by a linearly growing <scale factor>.
There is a literal qualification to the claim that $a$ must reach zero: the printed hypothesis alone does not exclude an earlier <curvature singularity> with positive limiting $a$. In arbitrary fixed units, take $k=0$ and $a(t)=1+\sqrt t$ for $t>0$, and define $\varrho,p$ using the two <Friedmann equations> and the <Friedmann acceleration equation>. Then $\dot a>0$, $\ddot a<0$, and $\varrho+3p=-3\ddot a/(4\pi Ga)>0$, but $a\to1$ while the <energy density> and <pressure> diverge as $t\downarrow0$. Thus the rigorous unconditional conclusion is the finite bound on a regular past interval; reaching zero additionally requires the stated continuation assumption. Nor does a zero of $a$ alone prove a <curvature singularity>, as the <Milne model> illustrates.
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