= Solution
Differentiation of the <Hubble parameter> gives $\dot H+H^2=\ddot a/a$. The <strong energy condition> and the acceleration equation therefore imply
$$
\boxed{\dot H+H^2\leq0.}
$$
Assume the present universe is expanding, $H_0>0$, and normalize the <scale factor> by $a(t_0)=1$. Since $\ddot a\leq0$, the past expanding branch has $\dot a(t)\geq\dot a(t_0)>0$. On this branch,
$$
\frac{d}{dt}\frac1H=-\frac{\dot H}{H^2}\geq1.
$$
Integrating from a past time $t$ to $t_0$ gives $H(t)^{-1}\leq H_0^{-1}+t-t_0$. Thus, where the denominator is positive,
$$
\boxed{H(t)\geq\frac1{H_0^{-1}+t-t_0}\quad(t\leq t_0).}
$$
The printed inequality needs this past-time qualification: for future times with $H>0$, the inequality reverses. Independently, a <concave function> lies below its tangent, so
$$
\boxed{a(t)\leq1+H_0(t-t_0).}
$$
Positivity of $a$ prevents the past branch extending before the zero of this tangent. If $t_B$ is the beginning of the expanding universe, its age satisfies $\boxed{t_0-t_B\leq H_0^{-1}}$. Without the normalization $a(t_0)=1$, the tangent bound instead reads $a(t)\leq a(t_0)[1+H_0(t-t_0)]$.
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