= Solution
With spatial flatness, the scalar as the only source, and $M_{\rm Pl}=(8\pi G)^{-1/2}$, the <Friedmann equation> is
$$
\boxed{H^2=\frac{\rho}{3M_{\rm Pl}^2}=\frac{\dot\phi^2/2+V}{3M_{\rm Pl}^2}.}
$$
The scalar equation gives $\dot\rho=\dot\phi(\ddot\phi+V_{,\phi})=-3H\dot\phi^2$, which is the energy-conservation equation $\dot\rho+3H(\rho+P)=0$. Differentiating the <Friedmann equation> on the expanding branch yields $6M_{\rm Pl}^2H\dot H=-3H\dot\phi^2$, so $\dot H=-\dot\phi^2/(2M_{\rm Pl}^2)$. Combining this with $\ddot a/a=H^2+\dot H$ gives the <Friedmann acceleration equation>:
$$
\boxed{\frac{\ddot a}{a}=\frac{V-\dot\phi^2}{3M_{\rm Pl}^2}=-\frac{\rho+3P}{6M_{\rm Pl}^2}.}
$$
Thus \b[<cosmic inflation> occurs exactly when $V(\phi)>\dot\phi^2$] in this canonical scalar model. Equivalently $P<-\rho/3$, or the <first Hubble slow-roll parameter> $\epsilon_H=-\dot H/H^2$ is smaller than one.
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